## peeking into relocation of function static in shared library

Here’s GUI (TUI) output of a function with a static variable access:

B+ |0x7ffff7616800 <st32>           test   %edi,%edi                                                                                                                     |
|0x7ffff7616802 <st32+2>         je     0x7ffff7616811 <st32+17>                                                                                                      |
|0x7ffff7616804 <st32+4>         mov    %edi,%eax                                                                                                                     |
|0x7ffff7616806 <st32+6>         bswap  %eax                                                                                                                          |
|0x7ffff7616808 <st32+8>         mov    %eax,0x200852(%rip)        # 0x7ffff7817060 <st32.yst32>                                                                        |
|0x7ffff761680e <st32+14>        mov    %edi,%eax                                                                                                                     |
|0x7ffff7616810 <st32+16>        retq                                                                                                                                 |
>|0x7ffff7616811 <st32+17>        mov    0x200849(%rip),%edi        # 0x7ffff7817060 <st32.yst32>                                                                        |
|0x7ffff7616817 <st32+23>        bswap  %edi                                                                                                                          |
|0x7ffff7616819 <st32+25>        mov    %edi,%eax                                                                                                                     |
|0x7ffff761681b <st32+27>        retq                                                                                                                                 |
|0x7ffff761681c <_fini>          sub    $0x8,%rsp | |0x7ffff7616820 <_fini+4> add$0x8,%rsp                                                                                                                     |
|0x7ffff7616824 <_fini+8>        retq                                                                                                                                 |
+---------------------------------------------------------------------------------------------------------------------------------------------------------------------+


The associated code is:

int st32( int v ) {
static int yst32 = 0x1a2b3c4d;

if ( v ) {
yst32 = v;
}

return yst32;
}


The object code dump (prior to relocation) just has zeros in the offset for the variable:

$objdump -d g.bs.o | grep -A12 '<st32>' 0000000000000050 <st32>: 50: 85 ff test %edi,%edi 52: 74 0d je 61 <st32+0x11> 54: 89 f8 mov %edi,%eax 56: 0f c8 bswap %eax 58: 89 05 00 00 00 00 mov %eax,0x0(%rip) # 5e <st32+0xe> 5e: 89 f8 mov %edi,%eax 60: c3 retq 61: 8b 3d 00 00 00 00 mov 0x0(%rip),%edi # 67 <st32+0x17> 67: 0f cf bswap %edi 69: 89 f8 mov %edi,%eax 6b: c3 retq  The linker has filled in the real offsets in question, and the dynamic loader has collaborated to put the data segment in the desired location. The observant reader may notice bwsap instructions in the listings above that don’t make sense for x86_64 code. That is because this code is compiled with an LLVM pass that performs byte swapping at load and store points, making it big endian in a limited fashion. The book Linkers and Loaders has some nice explanation of how relocation works, but I wanted to see the end result first hand in the debugger. It turned out that my naive expectation that the sum of$rip and the constant relocation factor is the address of the global variable (actually static in this case) is incorrect. Check that out in the debugger:

(gdb) p /x 0x200849+$rip$1 = 0x7ffff781705a

(gdb) x/10 $1 0x7ffff781705a <gy+26>: 0x22110000 0x2b1a4433 0x00004d3c 0x00000000 0x7ffff781706a: 0x00000000 0x00000000 0x00000000 0x30350000 0x7ffff781707a: 0x20333236 0x64655228  My magic value 0x1a2b3c4d looks like it is 6 bytes into the$rip + 0x200849 location that the disassembly appears to point to, and that is in fact the case:

(gdb) x/10 $1+6 0x7ffff7817060 <st32.yst32>: 0x4d3c2b1a 0x00000000 0x00000000 0x00000000 0x7ffff7817070 <y32>: 0x00000000 0x00000000 0x32363035 0x52282033 0x7ffff7817080: 0x48206465 0x34207461  My guess was the mysterious offset of 6 required to actually find this global address was the number of bytes in the MOV instruction, and sure enough that MOV is 6 bytes long: (gdb) disassemble /r Dump of assembler code for function st32: 0x00007ffff7616800 <+0>: 85 ff test %edi,%edi 0x00007ffff7616802 <+2>: 74 0d je 0x7ffff7616811 <st32+17> 0x00007ffff7616804 <+4>: 89 f8 mov %edi,%eax 0x00007ffff7616806 <+6>: 0f c8 bswap %eax 0x00007ffff7616808 <+8>: 89 05 52 08 20 00 mov %eax,0x200852(%rip) # 0x7ffff7817060 <st32.yst32> 0x00007ffff761680e <+14>: 89 f8 mov %edi,%eax 0x00007ffff7616810 <+16>: c3 retq => 0x00007ffff7616811 <+17>: 8b 3d 49 08 20 00 mov 0x200849(%rip),%edi # 0x7ffff7817060 <st32.yst32> 0x00007ffff7616817 <+23>: 0f cf bswap %edi 0x00007ffff7616819 <+25>: 89 f8 mov %edi,%eax 0x00007ffff761681b <+27>: c3 retq End of assembler dump.  So, it appears that the %rip reference in the disassembly is really the value of the instruction pointer after the instruction executes, which is curious. Note that this 4 byte relocation requires the shared library code segment and the shared library data segment be separated by no more than 4G. The linux dynamic loader has put all the segments back to back so that this is the case. This can be seen from /proc/PID/maps for the process: $ ps -ef | grep maindl
pjoot    17622 17582  0 10:50 pts/3    00:00:00 /home/pjoot/workspace/pass/global/maindl libglobtestbs.so

\$ grep libglob /proc/17622/maps
7ffff7616000-7ffff7617000 r-xp 00000000 fc:00 2492653                    /home/pjoot/workspace/pass/global/libglobtestbs.so
7ffff7617000-7ffff7816000 ---p 00001000 fc:00 2492653                    /home/pjoot/workspace/pass/global/libglobtestbs.so
7ffff7816000-7ffff7817000 r--p 00000000 fc:00 2492653                    /home/pjoot/workspace/pass/global/libglobtestbs.so
7ffff7817000-7ffff7818000 rw-p 00001000 fc:00 2492653                    /home/pjoot/workspace/pass/global/libglobtestbs.so


We’ve got a read-execute mmap region, where the code lies, and a read-write mmap region for the data. There’s a read-only segment which I presume is for read only global variables (my shared lib has one such variable and we have one page worth of space allocated for read only memory).

I wonder what the segment that has none of the read, write, nor execute permissions set is?

February 24, 2017 C/C++ development and debugging. No comments , ,

Fresh off the press:

I got this book to get some background on relocation of ELF globals, and was surprised to find a bit on z/OS (punch card compatible!) object format layout:

… an interesting bonus that’s topical.

## Omelets, a nice perk of working from home.

February 9, 2017 Food No comments , , ,

Today’s lunch was a three egg avocado, onion, red pepper, mushroom, cheese omelet, made with free range eggs from a little local Markham farm (16th just past York Durham line) .  I should have included a picture of the eggs before I cracked them, since two of them were an awesome blue, which the farm owner said is due to the Brazilian breed.

Prep included using Sofia’s awesome trick of separating the eggs, and pre-whipping the whites before a final recombination:

and the final delicious result:

I realize now that I forgot to add milk this time, which I’ve done in the past, but it all still worked out well.  I guess the milk is not really required.  As I tend to do with tortillas, I overstuffed this, and now I’m also overstuffed.

## Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper, from [1].

## Today

• Finish local vs global.
• Compositions of functions.
• Introduction to convex optimization problems.

## Continuing proof:

We want to prove that if

\begin{equation*}
\begin{aligned}
\end{aligned},
\end{equation*}

then $$\Bx^\conj$$ is a local optimum.

Proof:

Again, using Taylor approximation

\label{eqn:convexOptimizationLecture8:20}
F(\Bx^\conj + \Bv) = F(\Bx^\conj) + \lr{ \spacegrad F(\Bx^\conj)}^\T \Bv + \inv{2} \Bv^\T \spacegrad^2 F(\Bx^\conj) \Bv + o(\Norm{\Bv}^2)

The linear term is zero by assumption, whereas the Hessian term is given as $$> 0$$. Any direction that you move in, if your move is small enough, this is going uphill at a local optimum.

## Summarize:

For twice continuously differentiable functions, at a local optimum $$\Bx^\conj$$, then

\label{eqn:convexOptimizationLecture8:40}
\begin{aligned}
\end{aligned}

If, in addition, $$F$$ is convex, then $$\spacegrad F(\Bx^\conj) = 0$$ implies that $$\Bx^\conj$$ is a global optimum. i.e. for (unconstrained) convex functions, local and global optimums are equivalent.

• It is possible that a convex function does not have a global optimum. Examples are $$F(x) = e^x$$
(fig. 1)
, which has an $$\inf$$, but no lowest point.

fig. 1. Exponential has no global optimum.

• Our discussion has been for unconstrained functions. For constrained problems (next topic) is not not necessarily true that $$\spacegrad F(\Bx) = 0$$ implies that $$\Bx$$ is a global optimum, even for $$F$$ convex.

As an example of a constrained problem consider

\label{eqn:convexOptimizationLecture8:n}
\begin{aligned}
\min &2 x^2 + y^2 \\
x &\ge 3 \\
y &\ge 5.
\end{aligned}

The level sets of this objective function are plotted in fig. 2. The optimal point is at $$\Bx^\conj = (3,5)$$, where $$\spacegrad F \ne 0$$.

fig. 2. Constrained problem with optimum not at the zero gradient point.

## Projection

Given $$\Bx \in \mathbb{R}^n, \By \in \mathbb{R}^p$$, if $$h(\Bx,\By)$$ is convex in $$\Bx, \By$$, then

\label{eqn:convexOptimizationLecture8:60}
F(\Bx_0) = \inf_\By h(\Bx_0,\By)

is convex in $$\Bx$$, as sketched in fig. 3.

fig. 3. Epigraph of $$h$$ is a filled bowl.

The intuition here is that shining light on the (filled) “bowl”. That is, the image of $$\textrm{epi} h$$ on the $$\By = 0$$ screen which we will show is a convex set.

Proof:

Since $$h$$ is convex in $$\begin{bmatrix} \Bx \\ \By \end{bmatrix} \in \textrm{dom} h$$, then

\label{eqn:convexOptimizationLecture8:80}
\textrm{epi} h = \setlr{ (\Bx,\By,t) | t \ge h(\Bx,\By), \begin{bmatrix} \Bx \\ \By \end{bmatrix} \in \textrm{dom} h },

is a convex set.

We also have to show that the domain of $$F$$ is a convex set. To show this note that

\label{eqn:convexOptimizationLecture8:100}
\begin{aligned}
\textrm{dom} F
&= \setlr{ \Bx | \exists \By s.t. \begin{bmatrix} \Bx \\ \By \end{bmatrix} \in \textrm{dom} h } \\
&= \setlr{
\begin{bmatrix}
I_{n\times n} & 0_{n \times p}
\end{bmatrix}
\begin{bmatrix}
\Bx \\
\By
\end{bmatrix}
| \begin{bmatrix} \Bx \\ \By \end{bmatrix} \in \textrm{dom} h
}.
\end{aligned}

This is an affine map of a convex set. Therefore $$\textrm{dom} F$$ is a convex set.

\label{eqn:convexOptimizationLecture8:120}
\begin{aligned}
\textrm{epi} F
&=
\setlr{ \begin{bmatrix} \Bx \\ \By \end{bmatrix} | t \ge \inf h(\Bx,\By), \Bx \in \textrm{dom} F, \By: \begin{bmatrix} \Bx \\ \By \end{bmatrix} \in \textrm{dom} h } \\
&=
\setlr{
\begin{bmatrix}
I & 0 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
\Bx \\
\By \\
t
\end{bmatrix}
|
t \ge h(\Bx,\By), \begin{bmatrix} \Bx \\ \By \end{bmatrix} \in \textrm{dom} h
}.
\end{aligned}

### Example:

The function

\label{eqn:convexOptimizationLecture8:140}
F(\Bx) = \inf_{\By \in C} \Norm{ \Bx – \By },

over $$\Bx \in \mathbb{R}^n, \By \in C$$, ,is convex if $$C$$ is a convex set. Reason:

• $$\Bx – \By$$ is linear in $$(\Bx, \By)$$.
• $$\Norm{ \Bx – \By }$$ is a convex function if the domain is a convex set
• The domain is $$\mathbb{R}^n \times C$$. This will be a convex set if $$C$$ is.
• $$h(\Bx, \By) = \Norm{\Bx -\By}$$ is a convex function if $$\textrm{dom} h$$ is a convex set. By setting $$\textrm{dom} h = \mathbb{R}^n \times C$$, if $$C$$ is convex, $$\textrm{dom} h$$ is a convex set.
• $$F()$$

## Composition of functions

Consider

\label{eqn:convexOptimizationLecture8:160}
\begin{aligned}
F(\Bx) &= h(g(\Bx)) \\
\textrm{dom} F &= \setlr{ \Bx \in \textrm{dom} g | g(\Bx) \in \textrm{dom} h } \\
F &: \mathbb{R}^n \rightarrow \mathbb{R} \\
g &: \mathbb{R}^n \rightarrow \mathbb{R} \\
h &: \mathbb{R} \rightarrow \mathbb{R}.
\end{aligned}

Cases:

1. $$g$$ is convex, $$h$$ is convex and non-decreasing.
2. $$g$$ is convex, $$h$$ is convex and non-increasing.

Show for 1D case ( $$n = 1$$). Get to $$n > 1$$ by applying to all lines.

1. \label{eqn:convexOptimizationLecture8:180}
\begin{aligned}
F'(x) &= h'(g(x)) g'(x) \\
F”(x) &=
h”(g(x)) g'(x) g'(x)
+
h'(g(x)) g”(x) \\
&=
h”(g(x)) (g'(x))^2
+
h'(g(x)) g”(x) \\
&=
\lr{ \ge 0 } \cdot \lr{ \ge 0 }^2 + \lr{ \ge 0 } \cdot \lr{ \ge 0 },
\end{aligned}

since $$h$$ is respectively convex, and non-decreasing.

2. \label{eqn:convexOptimizationLecture8:180b}
\begin{aligned}
F'(x) =
\lr{ \ge 0 } \cdot \lr{ \ge 0 }^2 + \lr{ \le 0 } \cdot \lr{ \le 0 },
\end{aligned}

since $$h$$ is respectively convex, and non-increasing, and g is concave.

## Extending to multiple dimensions

\label{eqn:convexOptimizationLecture8:200}
\begin{aligned}
F(\Bx)
&= h(g(\Bx)) = h( g_1(\Bx), g_2(\Bx), \cdots g_k(\Bx) ) \\
g &: \mathbb{R}^n \rightarrow \mathbb{R} \\
h &: \mathbb{R}^k \rightarrow \mathbb{R}.
\end{aligned}

is convex if $$g_i$$ is convex for each $$i \in [1,k]$$ and $$h$$ is convex and non-decreasing in each argument.

Proof:

again assume $$n = 1$$, without loss of generality,

\label{eqn:convexOptimizationLecture8:220}
\begin{aligned}
g &: \mathbb{R} \rightarrow \mathbb{R}^k \\
h &: \mathbb{R}^k \rightarrow \mathbb{R} \\
\end{aligned}

\label{eqn:convexOptimizationLecture8:240}
F”(\Bx)
=
\begin{bmatrix}
g_1(\Bx) & g_2(\Bx) & \cdots & g_k(\Bx)
\end{bmatrix}
\begin{bmatrix}
g_1′(\Bx) \\ g_2′(\Bx) \\ \vdots \\ g_k'(\Bx)
\end{bmatrix}
+
\begin{bmatrix}
g_1”(\Bx) \\ g_2”(\Bx) \\ \vdots \\ g_k”(\Bx)
\end{bmatrix}

The Hessian is PSD.

### Example:

\label{eqn:convexOptimizationLecture8:260}
F(x) = \exp( g(x) ) = h( g(x) ),

where $$g$$ is convex is convex, and $$h(y) = e^y$$. This implies that $$F$$ is a convex function.

### Example:

\label{eqn:convexOptimizationLecture8:280}
F(x) = \inv{g(x)},

is convex if $$g(x)$$ is concave and positive. The most simple such example of such a function is $$h(x) = 1/x, \textrm{dom} h = \mathbb{R}_{++}$$, which is plotted in fig. 4.

fig. 4. Inverse function is convex over positive domain.

### Example:

\label{eqn:convexOptimizationLecture8:300}
F(x) = – \sum_{i = 1}^n \log( -F_i(x) )

is convex on $$\setlr{ x | F_i(x) < 0 \forall i }$$ if all $$F_i$$ are convex.

• Due to $$\textrm{dom} F$$, $$-F_i(x) > 0 \,\forall x \in \textrm{dom} F$$
• $$\log(x)$$ concave on $$\mathbb{R}_{++}$$ so $$-\log$$ convex also non-increasing (fig. 5).

fig. 5. Negative logarithm convex over positive domain.

\label{eqn:convexOptimizationLecture8:320}
F(x) = \sum h_i(x)

but
\label{eqn:convexOptimizationLecture8:340}
h_i(x) = -\log(-F_i(x)),

which is a convex and non-increasing function ($$-\log$$), of a convex function $$-F_i(x)$$. Each
$$h_i$$ is convex, so this is a sum of convex functions, and is therefore convex.

### Example:

Over $$\textrm{dom} F = S^n_{++}$$

\label{eqn:convexOptimizationLecture8:360}
F(X) = \log \det X^{-1}

To show that this is convex, check all lines in domain. A line in $$S^n_{++}$$ is a 1D family of matrices

\label{eqn:convexOptimizationLecture8:380}
\tilde{F}(t) = \log \det( \lr{X_0 + t H}^{-1} ),

where $$X_0 \in S^n_{++}, t \in \mathbb{R}, H \in S^n$$.

F9

For $$t$$ small enough,

\label{eqn:convexOptimizationLecture8:400}
X_0 + t H \in S^n_{++}

\label{eqn:convexOptimizationLecture8:420}
\begin{aligned}
\tilde{F}(t)
&= \log \det( \lr{X_0 + t H}^{-1} ) \\
&= \log \det\lr{ X_0^{-1/2} \lr{I + t X_0^{-1/2} H X_0^{-1/2} }^{-1} X_0^{-1/2} } \\
&= \log \det\lr{ X_0^{-1} \lr{I + t X_0^{-1/2} H X_0^{-1/2} }^{-1} } \\
&= \log \det X_0^{-1} + \log\det \lr{I + t X_0^{-1/2} H X_0^{-1/2} }^{-1} \\
&= \log \det X_0^{-1} – \log\det \lr{I + t X_0^{-1/2} H X_0^{-1/2} } \\
&= \log \det X_0^{-1} – \log\det \lr{I + t M }.
\end{aligned}

If $$\lambda_i$$ are eigenvalues of $$M$$, then $$1 + t \lambda_i$$ are eigenvalues of $$I + t M$$. i.e.:

\label{eqn:convexOptimizationLecture8:440}
\begin{aligned}
(I + t M) \Bv
&=
I \Bv + t \lambda_i \Bv \\
&=
(1 + t \lambda_i) \Bv.
\end{aligned}

This gives

\label{eqn:convexOptimizationLecture8:460}
\begin{aligned}
\tilde{F}(t)
&= \log \det X_0^{-1} – \log \prod_{i = 1}^n (1 + t \lambda_i) \\
&= \log \det X_0^{-1} – \sum_{i = 1}^n \log (1 + t \lambda_i)
\end{aligned}

• $$1 + t \lambda_i$$ is linear in $$t$$.
• $$-\log$$ is convex in its argument.
• sum of convex function is convex.

### Example:

\label{eqn:convexOptimizationLecture8:480}
F(X) = \lambda_\max(X),

is convex on $$\textrm{dom} F \in S^n$$

(a)
\label{eqn:convexOptimizationLecture8:500}
\lambda_{\max} (X) = \sup_{\Norm{\Bv}_2 \le 1} \Bv^\T X \Bv,

\label{eqn:convexOptimizationLecture8:520}
\begin{bmatrix}
\lambda_1 & & & \\
& \lambda_2 & & \\
& & \ddots & \\
& & & \lambda_n
\end{bmatrix}

Recall that a decomposition

\label{eqn:convexOptimizationLecture8:540}
\begin{aligned}
X &= Q \Lambda Q^\T \\
Q^\T Q = Q Q^\T = I
\end{aligned}

can be used for any $$X \in S^n$$.

(b)

Note that $$\Bv^\T X \Bv$$ is linear in $$X$$. This is a max of a number of linear (and convex) functions, so it is convex.

Last example:

(non-symmetric matrices)

\label{eqn:convexOptimizationLecture8:560}
F(X) = \sigma_\max(X),

is convex on $$\textrm{dom} F = \mathbb{R}^{m \times n}$$. Here

\label{eqn:convexOptimizationLecture8:580}
\sigma_\max(X) = \sup_{\Norm{\Bv}_2 = 1} \Norm{X \Bv}_2

This is called an operator norm of $$X$$. Using the SVD

\label{eqn:convexOptimizationLecture8:600}
\begin{aligned}
X &= U sectionigma V^\T \\
U &= \mathbb{R}^{m \times r} \\
sectionigma &\in \mathrm{diag} \in \mathbb{R}{ r \times r } \\
V^T &\in \mathbb{R}^{r \times n}.
\end{aligned}

Have

\label{eqn:convexOptimizationLecture8:620}
\Norm{X \Bv}_2^2
=
\Norm{ U sectionigma V^\T \Bv }_2^2
=
\Bv^\T V sectionigma U^\T U sectionigma V^\T \Bv
=
\Bv^\T V sectionigma sectionigma V^\T \Bv
=
\Bv^\T V sectionigma^2 V^\T \Bv
=
\tilde{\Bv}^\T sectionigma^2 \tilde{\Bv},

where $$\tilde{\Bv} = \Bv^\T V$$, so

\label{eqn:convexOptimizationLecture8:640}
\Norm{X \Bv}_2^2
=
\sum_{i = 1}^r \sigma_i^2 \Norm{\tilde{\Bv}}
\le \sigma_\max^2 \Norm{\tilde{\Bv}}^2,

or
\label{eqn:convexOptimizationLecture8:660}
\Norm{X \Bv}_2
\le \sqrt{ \sigma_\max^2 } \Norm{\tilde{\Bv}}
\le
\sigma_\max.

Set $$\Bv$$ to the right singular value of $$X$$ to get equality.

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.

## Posted notes for Electromagnetic Theory (ECE1228H), taught by Prof. M. Mojahedi, fall 2016

February 3, 2017 math and physics play No comments ,

I’ve now posted redacted notes for the Electromagnetic Theory (ECE1228H) course I took last fall, taught by Prof. M. Mojahedi.  This course covered a subset of the following:

• Maxwell’s equations
• constitutive relations and boundary conditions
• wave polarization.
• Field representations: potentials
• Green’s functions and integral equations.
• Theorems and concepts: duality, uniqueness, images, equivalence, reciprocity and Babinet’s principles.
• Plane cylindrical and spherical waves and waveguides.

These notes are fairly compact, only 183 pages, with the full version weighing in at 256 pages.

As always, feel free to contact me for the complete version (i.e. including my problem set solutions) if you interested, but not asking because you are taking or planning to take this course.

## ECE1505H Convex Optimization. Lecture 7: Examples of convex and concave functions, local and global minimums. Taught by Prof. Stark Draper

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper, from [1].

### Today

• Local and global optimality
• Compositions of functions
• Examples

### Example:

\label{eqn:convexOptimizationLecture7:20}
\begin{aligned}
F(x) &= x^2 \\
F”(x) &= 2 > 0
\end{aligned}

strictly convex.

### Example:

\label{eqn:convexOptimizationLecture7:40}
\begin{aligned}
F(x) &= x^3 \\
F”(x) &= 6 x.
\end{aligned}

Not always non-negative, so not convex. However $$x^3$$ is convex on $$\textrm{dom} F = \mathbb{R}_{+}$$.

### Example:

\label{eqn:convexOptimizationLecture7:60}
\begin{aligned}
F(x) &= x^\alpha \\
F'(x) &= \alpha x^{\alpha-1} \\
F”(x) &= \alpha(\alpha-1) x^{\alpha-2}.
\end{aligned}

fig. 1. Powers of x.

This is convex on $$\mathbb{R}_{+}$$, if $$\alpha \ge 1$$, or $$\alpha \le 0$$.

### Example:

\label{eqn:convexOptimizationLecture7:80}
\begin{aligned}
F(x) &= \log x \\
F'(x) &= \inv{x} \\
F”(x) &= -\inv{x^2} \le 0
\end{aligned}

This is concave.

### Example:

\label{eqn:convexOptimizationLecture7:100}
\begin{aligned}
F(x) &= x\log x \\
F'(x) &= \log x + x \inv{x} = 1 + \log x \\
F”(x) &= \inv{x}
\end{aligned}

This is strictly convex on
$$\mathbb{R}_{++}$$, where
$$F”(x) \ge 0$$.

### Example:

\label{eqn:convexOptimizationLecture7:120}
\begin{aligned}
F(x) &= e^{\alpha x} \\
F'(x) &= \alpha e^{\alpha x} \\
F”(x) &= \alpha^2 e^{\alpha x} \ge 0
\end{aligned}

fig. 2. Exponential.

Such functions are plotted in fig. 2, and are convex function for all $$\alpha$$.

### Example:

For symmetric $$P \in S^n$$

\label{eqn:convexOptimizationLecture7:140}
\begin{aligned}
F(\Bx) &= \Bx^\T P \Bx + 2 \Bq^\T \Bx + r \\
\spacegrad F &= (P + P^\T) \Bx + 2 \Bq = 2 P \Bx + 2 \Bq \\
\end{aligned}

This is convex(concave) if $$P \ge 0$$ ($$P \le 0$$).

### Example:

\label{eqn:convexOptimizationLecture7:780}
F(x, y) = x^2 + y^2 + 3 x y,

that is neither convex nor concave is plotted in fig 3.

fig 3. Function with saddle point (3d and contours)

This function can be put in matrix form

\label{eqn:convexOptimizationLecture7:160}
F(x, y) = x^2 + y^2 + 3 x y
=
\begin{bmatrix}
x & y
\end{bmatrix}
\begin{bmatrix}
1 & 1.5 \\
1.5 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix},

and has the Hessian

\label{eqn:convexOptimizationLecture7:180}
\begin{aligned}
&=
\begin{bmatrix}
\partial_{xx} F & \partial_{xy} F \\
\partial_{yx} F & \partial_{yy} F \\
\end{bmatrix} \\
&=
\begin{bmatrix}
2 & 3 \\
3 & 2
\end{bmatrix} \\
&= 2 P.
\end{aligned}

From the plot we know that this is not PSD, but this can be confirmed by checking the eigenvalues

\label{eqn:convexOptimizationLecture7:200}
\begin{aligned}
0
&=
\det ( P – \lambda I ) \\
&=
(1 – \lambda)^2 – 1.5^2,
\end{aligned}

which has solutions

\label{eqn:convexOptimizationLecture7:220}
\lambda = 1 \pm \frac{3}{2} = \frac{3}{2}, -\frac{1}{2}.

This is not PSD nor negative semi-definite, because it has one positive and one negative eigenvalues. This is neither convex nor concave.

Along $$y = -x$$,

\label{eqn:convexOptimizationLecture7:240}
\begin{aligned}
F(x,y)
&=
F(x,-x) \\
&=
2 x^2 – 3 x^2 \\
&=
– x^2,
\end{aligned}

so it is concave along this line. Along $$y = x$$

\label{eqn:convexOptimizationLecture7:260}
\begin{aligned}
F(x,y)
&=
F(x,x) \\
&=
2 x^2 + 3 x^2 \\
&=
5 x^2,
\end{aligned}

so it is convex along this line.

### Example:

\label{eqn:convexOptimizationLecture7:280}
F(\Bx) = \sqrt{ x_1 x_2 },

on $$\textrm{dom} F = \setlr{ x_1 \ge 0, x_2 \ge 0 }$$

For the Hessian
\label{eqn:convexOptimizationLecture7:300}
\begin{aligned}
\PD{x_1}{F} &= \frac{1}{2} x_1^{-1/2} x_2^{1/2} \\
\PD{x_2}{F} &= \frac{1}{2} x_2^{-1/2} x_1^{1/2}
\end{aligned}

The Hessian components are

\label{eqn:convexOptimizationLecture7:320}
\begin{aligned}
\PD{x_1}{} \PD{x_1}{F} &= -\frac{1}{4} x_1^{-3/2} x_2^{1/2} \\
\PD{x_1}{} \PD{x_2}{F} &= \frac{1}{4} x_2^{-1/2} x_1^{-1/2} \\
\PD{x_2}{} \PD{x_1}{F} &= \frac{1}{4} x_1^{-1/2} x_2^{-1/2} \\
\PD{x_2}{} \PD{x_2}{F} &= -\frac{1}{4} x_2^{-3/2} x_1^{1/2}
\end{aligned}

or
\label{eqn:convexOptimizationLecture7:340}
=
-\frac{\sqrt{x_1 x_2}}{4}
\begin{bmatrix}
\inv{x_1^2} & -\inv{x_1 x_2} \\
-\inv{x_1 x_2} & \inv{x_2^2}
\end{bmatrix}.

Checking this for PSD against $$\Bv = (v_1, v_2)$$, we have
\label{eqn:convexOptimizationLecture7:360}
\begin{aligned}
\begin{bmatrix}
v_1 & v_2
\end{bmatrix}
\begin{bmatrix}
\inv{x_1^2} & -\inv{x_1 x_2} \\
-\inv{x_1 x_2} & \inv{x_2^2}
\end{bmatrix}
\begin{bmatrix}
v_1 \\ v_2
\end{bmatrix}
&=
\begin{bmatrix}
v_1 & v_2
\end{bmatrix}
\begin{bmatrix}
\inv{x_1^2} v_1 -\inv{x_1 x_2} v_2 \\
-\inv{x_1 x_2} v_1 + \inv{x_2^2} v_2
\end{bmatrix} \\
&=
\lr{ \inv{x_1^2} v_1 -\inv{x_1 x_2} v_2 } v_1 +
\lr{ -\inv{x_1 x_2} v_1 + \inv{x_2^2} v_2 } v_2
\\
&=
\inv{x_1^2} v_1^2
+ \inv{x_2^2} v_2^2
-2 \inv{x_1 x_2} v_1 v_2 \\
&=
\lr{
\frac{v_1}{x_1}
-\frac{v_2}{x_2}
}^2 \\
&\ge 0,
\end{aligned}

so $$\spacegrad^2 F \le 0$$. This is a negative semi-definite function (concave). Observe that this check required checking PSD for all values of $$\Bx$$.

This is an example of a more general result

\label{eqn:convexOptimizationLecture7:380}
F(x) = \lr{ \prod_{i = 1}^n x_i }^{1/n},

which is concave (prove on homework).

### Summary.

If $$F$$ is differentiable in \R{n}, then check the curvature of the function along all lines. i.e. At all locations and in all directions.

If the Hessian is PSD at all $$\Bx \in \textrm{dom} F$$, that is

\label{eqn:convexOptimizationLecture7:400}
\spacegrad^2 F \ge 0 \, \forall \Bx \in \textrm{dom} F,

then the function is convex.

### Example:

Over $$\textrm{dom} F = \mathbb{R}^n$$

\label{eqn:convexOptimizationLecture7:420}
F(\Bx) = \max_{i = 1}^n x_i

i.e.
\label{eqn:convexOptimizationLecture7:440}
\begin{aligned}
F((1,2) &= 2 \\
F((3,-1) &= 3
\end{aligned}

### Example:

\label{eqn:convexOptimizationLecture7:460}
F(\Bx) = \max_{i = 1}^n F_i(\Bx),

where

\label{eqn:convexOptimizationLecture7:480}
F_i(\Bx)
=
… ?

max of a set of convex functions is a convex function.

### Example:

\label{eqn:convexOptimizationLecture7:500}
F(x) =
x_{[1]} +
x_{[2]} +
x_{[3]}

where

$$x_{[k]}$$ is the k-th largest number in the list

Write

\label{eqn:convexOptimizationLecture7:520}
F(x) = \max x_i + x_j + x_k

\label{eqn:convexOptimizationLecture7:540}
(i,j,k) \in \binom{n}{3}

### Example:

For $$\Ba \in \mathbb{R}^n$$ and $$b_i \in \mathbb{R}$$

\label{eqn:convexOptimizationLecture7:560}
\begin{aligned}
F(\Bx)
&= \sum_{i = 1}^n \log( b_i – \Ba^\T \Bx )^{-1} \\
&= -\sum_{i = 1}^n \log( b_i – \Ba^\T \Bx )
\end{aligned}

This $$b_i – \Ba^\T \Bx$$ is an affine function of $$\Bx$$ so it doesn’t affect convexity.

Since $$\log$$ is concave, $$-\log$$ is convex. Convex functions of affine function of $$\Bx$$ is convex function of $$\Bx$$.

### Example:

\label{eqn:convexOptimizationLecture7:580}
F(\Bx) = \sup_{\By \in C} \Norm{ \Bx – \By }

fig. 3. Max length function

Here $$C \subseteq \mathbb{R}^n$$ is not necessarily convex. We are using $$\sup$$ here because the set $$C$$ may be open. This function is the length of the line from $$\Bx$$ to the point in $$C$$ that is furthest from $$\Bx$$.

• $$\Bx – \By$$ is linear in $$\Bx$$
• $$g_\By(\Bx) = \Norm{\Bx – \By}$$ is convex in $$\Bx$$ since norms are convex functions.
• $$F(\Bx) = \sup_{\By \in C} \Norm{ \Bx – \By }$$. Each $$\By$$ index is a convex function. Taking max of those.

### Example:

\label{eqn:convexOptimizationLecture7:600}
F(\Bx) = \inf_{\By \in C} \Norm{ \Bx – \By }.

Min and max of two convex functions are plotted in fig. 4.

fig. 4. Min and max

The max is observed to be convex, whereas the min is not necessarily so.

\label{eqn:convexOptimizationLecture7:800}
F(\Bz) = F(\theta \Bx + (1-\theta) \By) \ge \theta F(\Bx) + (1-\theta)F(\By).

This is not necessarily convex for all sets $$C \subseteq \mathbb{R}^n$$, because the $$\inf$$ of a bunch of convex function is not necessarily convex. However, if $$C$$ is convex, then $$F(\Bx)$$ is convex.

### Consequences of convexity for differentiable functions

• Think about unconstrained functions $$\textrm{dom} F = \mathbb{R}^n$$.
• By first order condition $$F$$ is convex iff the domain is convex and
\label{eqn:convexOptimizationLecture7:620}
F(\Bx) \ge \lr{ \spacegrad F(\Bx)}^\T (\By – \Bx) \, \forall \Bx, \By \in \textrm{dom} F.

If $$F$$ is convex and one can find an $$\Bx^\conj \in \textrm{dom} F$$ such that

\label{eqn:convexOptimizationLecture7:640}

then

\label{eqn:convexOptimizationLecture7:660}
F(\By) \ge F(\Bx^\conj) \, \forall \By \in \textrm{dom} F.

If you can find the point where the gradient is zero (which can’t always be found), then $$\Bx^\conj$$ is a global minimum of $$F$$.

Conversely, if $$\Bx^\conj$$ is a global minimizer of $$F$$, then $$\spacegrad F(\Bx^\conj) = 0$$ must hold. If that were not the case, then you would be able to find a direction to move downhill, contracting the optimality of $$\Bx^\conj$$.

### Local vs Global optimum

fig. 6. Global and local minimums

Definition: Local optimum
$$\Bx^\conj$$ is a local optimum of $$F$$ if $$\exists \epsilon > 0$$ such that $$\forall \Bx$$, $$\Norm{\Bx – \Bx^\conj} < \epsilon$$, we have

\begin{equation*}
F(\Bx^\conj) \le F(\Bx)
\end{equation*}

fig. 5. min length function

Theorem:
Suppose $$F$$ is twice continuously differentiable (not necessarily convex)

• If $$\Bx^\conj$$ is a local optimum then\begin{equation*}
\begin{aligned}
\end{aligned}
\end{equation*}
• If
\begin{equation*}
\begin{aligned}
\end{aligned},
\end{equation*}then $$\Bx^\conj$$ is a local optimum.

Proof:

• Let $$\Bx^\conj$$ be a local optimum. Pick any $$\Bv \in \mathbb{R}^n$$.\label{eqn:convexOptimizationLecture7:720}
\lim_{t \rightarrow 0} \frac{ F(\Bx^\conj + t \Bv) – F(\Bx^\conj)}{t}
= \lr{ \spacegrad F(\Bx^\conj) }^\T \Bv
\ge 0.

Here the fraction is $$\ge 0$$ since $$\Bx^\conj$$ is a local optimum.

Since the choice of $$\Bv$$ is arbitrary, the only case that you can ensure that $$\ge 0, \forall \Bv$$ is

\label{eqn:convexOptimizationLecture7:740}

( or else could pick $$\Bv = -\spacegrad F(\Bx^\conj)$$.

This means that $$\spacegrad F(\Bx^\conj) = 0$$ if $$\Bx^\conj$$ is a local optimum.

Consider the 2nd order derivative

\label{eqn:convexOptimizationLecture7:760}
\begin{aligned}
\lim_{t \rightarrow 0} \frac{ F(\Bx^\conj + t \Bv) – F(\Bx^\conj)}{t^2}
&=
\lim_{t \rightarrow 0} \inv{t^2}
\lr{
F(\Bx^\conj) + t \lr{ \spacegrad F(\Bx^\conj) }^\T \Bv + \inv{2} t^2 \Bv^\T \spacegrad^2 F(\Bx^\conj) \Bv + O(t^3)
– F(\Bx^\conj)
} \\
&=
\inv{2} \Bv^\T \spacegrad^2 F(\Bx^\conj) \Bv \\
&\ge 0.
\end{aligned}

Here the $$\ge$$ condition also comes from the fraction, based on the optimiality of $$\Bx^\conj$$. This is true for all choice of $$\Bv$$, thus $$\spacegrad^2 F(\Bx^\conj)$$.

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.

## ECE1505H Convex Optimization. Lecture 6: First and second order conditions. Taught by Prof.\ Stark Draper

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper, from [1].

### Today

• First and second order conditions for convexity of differentiable functions.
• Consequences of convexity: local and global optimality.
• Properties.

### Quasi-convex

$$F_1$$ and $$F_2$$ convex implies $$\max( F_1, F_2)$$ convex.

fig. 1. Min and Max

Note that $$\min(F_1, F_2)$$ is NOT convex.

If $$F : \mathbb{R}^n \rightarrow \mathbb{R}$$ is convex, then $$F( \Bx_0 + t \Bv )$$ is convex in $$t\,\forall t \in \mathbb{R}, \Bx_0 \in \mathbb{R}^n, \Bv \in \mathbb{R}^n$$, provided $$\Bx_0 + t \Bv \in \textrm{dom} F$$.

Idea: Restrict to a line (line segment) in $$\textrm{dom} F$$. Take a cross section or slice through $$F$$ alone the line. If the result is a 1D convex function for all slices, then $$F$$ is convex.

This is nice since it allows for checking for convexity, and is also nice numerically. Attempting to test a given data set for non-convexity with some random lines can help disprove convexity. However, to show that $$F$$ is convex it is required to test all possible slices (which isn’t possible numerically, but is in some circumstances possible analytically).

### Differentiable (convex) functions

Definition: First order condition.

If

\begin{equation*}
F : \mathbb{R}^n \rightarrow \mathbb{R}
\end{equation*}

is differentiable, then $$F$$ is convex iff $$\textrm{dom} F$$ is a convex set and $$\forall \Bx, \Bx_0 \in \textrm{dom} F$$

\begin{equation*}
F(\Bx) \ge F(\Bx_0) + \lr{\spacegrad F(\Bx_0)}^\T (\Bx – \Bx_0).
\end{equation*}

This is the first order Taylor expansion. If $$n = 1$$, this is $$F(x) \ge F(x_0) + F'(x_0) ( x – x_0)$$.

The first order condition says a convex function \underline{always} lies above its first order approximation, as sketched in fig. 3.

fig. 2. First order approximation lies below convex function

When differentiable, the supporting plane is the tangent plane.

Definition: Second order condition

If $$F : \mathbb{R}^n \rightarrow \mathbb{R}$$ is twice differentiable, then $$F$$ is convex iff $$\textrm{dom} F$$ is a convex set and $$\spacegrad^2 F(\Bx) \ge 0 \,\forall \Bx \in \textrm{dom} F$$.

The Hessian is always symmetric, but is not necessarily positive. Recall that the Hessian is the matrix of the second order partials $$(\spacegrad F)_{ij} = \partial^2 F/(\partial x_i \partial x_j)$$.

The scalar case is $$F”(x) \ge 0 \, \forall x \in \textrm{dom} F$$.

An implication is that if $$F$$ is convex, then $$F(x) \ge F(x_0) + F'(x_0) (x – x_0) \,\forall x, x_0 \in \textrm{dom} F$$

Since $$F$$ is convex, $$\textrm{dom} F$$ is convex.

Consider any 2 points $$x, y \in \textrm{dom} F$$, and $$\theta \in [0,1]$$. Define

\label{eqn:convexOptimizationLecture6:60}
z = (1-\theta) x + \theta y \in \textrm{dom} F,

then since $$\textrm{dom} F$$ is convex

\label{eqn:convexOptimizationLecture6:80}
F(z) =
F( (1-\theta) x + \theta y )
\le
(1-\theta) F(x) + \theta F(y )

Reordering

\label{eqn:convexOptimizationLecture6:220}
\theta F(x) \ge
\theta F(x) + F(z) – F(x),

or
\label{eqn:convexOptimizationLecture6:100}
F(y) \ge
F(x) + \frac{F(x + \theta(y-x)) – F(x)}{\theta},

which is, in the limit,

\label{eqn:convexOptimizationLecture6:120}
F(y) \ge
F(x) + F'(x) (y – x),

completing one direction of the proof.

To prove the other direction, showing that

\label{eqn:convexOptimizationLecture6:140}
F(x) \ge F(x_0) + F'(x_0) (x – x_0),

implies that $$F$$ is convex. Take any $$x, y \in \textrm{dom} F$$ and any $$\theta \in [0,1]$$. Define

\label{eqn:convexOptimizationLecture6:160}
z = \theta x + (1 -\theta) y,

which is in $$\textrm{dom} F$$ by assumption. We want to show that

\label{eqn:convexOptimizationLecture6:180}
F(z) \le \theta F(x) + (1-\theta) F(y).

By assumption

1. $$F(x) \ge F(z) + F'(z) (x – z)$$
2. $$F(y) \ge F(z) + F'(z) (y – z)$$

Compute

\label{eqn:convexOptimizationLecture6:200}
\begin{aligned}
\theta F(x) + (1-\theta) F(y)
&\ge
\theta \lr{ F(z) + F'(z) (x – z) }
+ (1-\theta) \lr{ F(z) + F'(z) (y – z) } \\
&=
F(z) + F'(z) \lr{ \theta( x – z) + (1-\theta) (y-z) } \\
&=
F(z) + F'(z) \lr{ \theta x + (1-\theta) y – \theta z – (1 -\theta) z } \\
&=
F(z) + F'(z) \lr{ \theta x + (1-\theta) y – z} \\
&=
F(z) + F'(z) \lr{ z – z} \\
&= F(z).
\end{aligned}

### Proof of the 2nd order case for $$n = 1$$

Want to prove that if

\label{eqn:convexOptimizationLecture6:240}
F : \mathbb{R} \rightarrow \mathbb{R}

is a convex function, then $$F”(x) \ge 0 \,\forall x \in \textrm{dom} F$$.

By the first order conditions $$\forall x \ne y \in \textrm{dom} F$$

\label{eqn:convexOptimizationLecture6:260}
\begin{aligned}
F(y) &\ge F(x) + F'(x) (y – x)
F(x) &\ge F(y) + F'(y) (x – y)
\end{aligned}

Can combine and get

\label{eqn:convexOptimizationLecture6:280}
F'(x) (y-x) \le F(y) – F(x) \le F'(y)(y-x)

Subtract the two derivative terms for

\label{eqn:convexOptimizationLecture6:340}
\frac{(F'(y) – F'(x))(y – x)}{(y – x)^2} \ge 0,

or
\label{eqn:convexOptimizationLecture6:300}
\frac{F'(y) – F'(x)}{y – x} \ge 0.

In the limit as $$y \rightarrow x$$, this is
\label{eqn:convexOptimizationLecture6:320}
\boxed{
F”(x) \ge 0 \,\forall x \in \textrm{dom} F.
}

Now prove the reverse condition:

If $$F”(x) \ge 0 \,\forall x \in \textrm{dom} F \subseteq \mathbb{R}$$, implies that $$F : \mathbb{R} \rightarrow \mathbb{R}$$ is convex.

Note that if $$F”(x) \ge 0$$, then $$F'(x)$$ is non-decreasing in $$x$$.

i.e. If $$x < y$$, where $$x, y \in \textrm{dom} F$$, then

\label{eqn:convexOptimizationLecture6:360}
F'(x) \le F'(y).

Consider any $$x,y \in \textrm{dom} F$$ such that $$x < y$$, where

\label{eqn:convexOptimizationLecture6:380}
F(y) – F(x) = \int_x^y F'(t) dt \ge F'(x) \int_x^y 1 dt = F'(x) (y-x).

This tells us that

\label{eqn:convexOptimizationLecture6:400}
F(y) \ge F(x) + F'(x)(y – x),

which is the first order condition. Similarly consider any $$x,y \in \textrm{dom} F$$ such that $$x < y$$, where

\label{eqn:convexOptimizationLecture6:420}
F(y) – F(x) = \int_x^y F'(t) dt \le F'(y) \int_x^y 1 dt = F'(y) (y-x).

This tells us that

\label{eqn:convexOptimizationLecture6:440}
F(x) \ge F(y) + F'(y)(x – y).

### Vector proof:

$$F$$ is convex iff $$F(\Bx + t \Bv)$$ is convex $$\forall \Bx,\Bv \in \mathbb{R}^n, t \in \mathbb{R}$$, keeping $$\Bx + t \Bv \in \textrm{dom} F$$.

Let
\label{eqn:convexOptimizationLecture6:460}
h(t ; \Bx, \Bv) = F(\Bx + t \Bv)

then $$h(t)$$ satisfies scalar first and second order conditions for all $$\Bx, \Bv$$.

\label{eqn:convexOptimizationLecture6:480}
h(t) = F(\Bx + t \Bv) = F(g(t)),

where $$g(t) = \Bx + t \Bv$$, where

\label{eqn:convexOptimizationLecture6:500}
\begin{aligned}
F &: \mathbb{R}^n \rightarrow \mathbb{R} \\
g &: \mathbb{R} \rightarrow \mathbb{R}^n.
\end{aligned}

This is expressing $$h(t)$$ as a composition of two functions. By the first order condition for scalar functions we know that

\label{eqn:convexOptimizationLecture6:520}
h(t) \ge h(0) + h'(0) t.

Note that

\label{eqn:convexOptimizationLecture6:540}
h(0) = \evalbar{F(\Bx + t \Bv)}{t = 0} = F(\Bx).

Let’s figure out what $$h'(0)$$ is. Recall hat for any $$\tilde{F} : \mathbb{R}^n \rightarrow \mathbb{R}^m$$

\label{eqn:convexOptimizationLecture6:560}
D \tilde{F} \in \mathbb{R}^{m \times n},

and
\label{eqn:convexOptimizationLecture6:580}
{D \tilde{F}(\Bx)}_{ij} = \PD{x_j}{\tilde{F_i}(\Bx)}

This is one function per row, for $$i \in [1,m], j \in [1,n]$$. This gives

\label{eqn:convexOptimizationLecture6:600}
\begin{aligned}
\frac{d}{dt} F(\Bx + \Bv t)
&=
\frac{d}{dt} F( g(t) ) \\
&=
\frac{d}{dt} h(t) \\
&= D h(t) \\
&= D F(g(t)) \cdot D g(t)
\end{aligned}

The first matrix is in $$\mathbb{R}^{1\times n}$$ whereas the second is in $$\mathbb{R}^{n\times 1}$$, since $$F : \mathbb{R}^n \rightarrow \mathbb{R}$$ and $$g : \mathbb{R} \rightarrow \mathbb{R}^n$$. This gives

\label{eqn:convexOptimizationLecture6:620}
\frac{d}{dt} F(\Bx + \Bv t)
= \evalbar{D F(\tilde{\Bx})}{\tilde{\Bx} = g(t)} \cdot D g(t).

That first matrix is

\label{eqn:convexOptimizationLecture6:640}
\begin{aligned}
\evalbar{D F(\tilde{\Bx})}{\tilde{\Bx} = g(t)}
&=
\evalbar{
\lr{\begin{bmatrix}
\PD{\tilde{x}_1}{ F(\tilde{\Bx})} &
\PD{\tilde{x}_2}{ F(\tilde{\Bx})} & \cdots
\PD{\tilde{x}_n}{ F(\tilde{\Bx})}
\end{bmatrix}
}}{ \tilde{\Bx} = g(t) = \Bx + t \Bv } \\
&=
\evalbar{
}{
\tilde{\Bx} = g(t)
} \\
=
\end{aligned}

The second Jacobian is

\label{eqn:convexOptimizationLecture6:660}
D g(t)
=
D
\begin{bmatrix}
g_1(t) \\
g_2(t) \\
\vdots \\
g_n(t) \\
\end{bmatrix}
=
D
\begin{bmatrix}
x_1 + t v_1 \\
x_2 + t v_2 \\
\vdots \\
x_n + t v_n \\
\end{bmatrix}
=
\begin{bmatrix}
v_1 \\
v_1 \\
\vdots \\
v_n \\
\end{bmatrix}
=
\Bv.

so

\label{eqn:convexOptimizationLecture6:680}
h'(t) = D h(t) = \lr{ \spacegrad F(g(t))}^\T \Bv,

and
\label{eqn:convexOptimizationLecture6:700}
h'(0) = \lr{ \spacegrad F(g(0))}^\T \Bv
=

Finally

\label{eqn:convexOptimizationLecture6:720}
\begin{aligned}
F(\Bx + t \Bv)
&\ge h(0) + h'(0) t \\
&= F(\Bx) + \lr{ \spacegrad F(\Bx) }^\T (t \Bv) \\
&= F(\Bx) + \innerprod{ \spacegrad F(\Bx) }{ t \Bv}.
\end{aligned}

Which is true for all $$\Bx, \Bx + t \Bv \in \textrm{dom} F$$. Note that the quantity $$t \Bv$$ is a shift.

### Epigraph

Recall that if $$(\Bx, t) \in \textrm{epi} F$$ then $$t \ge F(\Bx)$$.

\label{eqn:convexOptimizationLecture6:740}
t \ge F(\Bx) \ge F(\Bx_0) + \lr{\spacegrad F(\Bx_0) }^\T (\Bx – \Bx_0),

or

\label{eqn:convexOptimizationLecture6:760}
0 \ge
-(t – F(\Bx_0)) + \lr{\spacegrad F(\Bx_0) }^\T (\Bx – \Bx_0),

In block matrix form

\label{eqn:convexOptimizationLecture6:780}
0 \ge
\begin{bmatrix}
\lr{ \spacegrad F(\Bx_0) }^\T & -1
\end{bmatrix}
\begin{bmatrix}
\Bx – \Bx_0 \\
t – F(\Bx_0)
\end{bmatrix}

With $$\Bw = \begin{bmatrix} \lr{ \spacegrad F(\Bx_0) }^\T & -1 \end{bmatrix}$$, the geometry of the epigraph relation to the half plane is sketched in fig. 3.

fig. 3. Half planes and epigraph.

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.