\(\def\Real{\mathbb{R}}
\def\Comp{\mathbb{C}}
\def\Rat{\mathbb{Q}}
\def\Field{\mathbb{F}}
\)
Sample problems
 Find Fenchel dual (conjugate, or \(\check{f}(p):=\sup_x pxf(x)\)) for
 \(f(x)=x1+x/2\);
Answer:
\[
\check{f}(p)=p1/2 \mathrm{\ if}\ 1/2\leq p\leq 3/2; +\infty \mathrm{\ otherwise}.
\]  f\((x)=x^2/2+1/2 \ \mathrm{for} \ x\leq 1; x \ \mathrm{otherwise}\).
Answer:
\[
\check{f}(p)=p^2/21/2 \mathrm{\ if}\ 1\leq p\leq 1; +\infty \mathrm{\ otherwise}.
\]
 \(f(x)=x1+x/2\);

Is the subset of \(\Real^3\) given by
\[
a\gt 0; a^2b^2\gt 0; a^3a(b^2+c^2)\gt 0
\]
convex?Hint: when \(
\left(
\begin{array}{ccc}
a&b&0\\
b&a&c\\
0&c&a\\
\end{array}
\right)
\) is positive semidefinite?
Answer: yes, the inequalities are the Sylvester condition for the matrix to be positive definite;
convex combination of positive definite matrices is positive definite again. 
Find the conditions on vector \((a,b)\) to be a subdifferential for
\[
f(x,y)=\max(e^x,e^y,e^{xy})
\]
at the points \((0,0), (2,1),(2,1)\).Answer: for \((0,0)\), \((a,b)\) is in the convex hull of the vectors \((0,1),(1,0),(1,1)\); for \((2,1)\), \((a,b)\) is in the convex hull of the vectors \((0,e),(e,e)\); for \((2,1)\), \((a,b)=(e^2,0)\).

Consider the problem
\[
x=(x_1,\ldots,x_d); f(x)\to \max,\ \mathrm{subject\ to}\ \sum_i p_i x_i\leq b, x_i\gt 0, i=1,\ldots, d,
\]
where \(f(x)=\prod x_i^{a_i}, a_i\gt 0\) is the CobbDouglas utility function (a standard microeconomic setting, where \(x_i\) are quantities of \(i\)th good, \(p_i\) are corresponding prices, and \(b\) is the budget constraint). Is the optimum unique?
Answer: Yes, it is given by
\[
x_i=\frac{ba_i}{ap_i},
\]
where \(a=\sum_i a_i\).  If \(\{x_i(b)\}_{i=1,\ldots,d}\) is the optimal solution, find \({dx_i}/{db}\).
Answer: \[
{dx_i}/{db}=\frac{a_i}{ap_i},
\]  Can you find a concave \(f\) such that for some \(i\), \({dx_i}/{db}\lt 0\)?
Answer: take, for example,
\[
f(x_1,x_2)=\min(x_1+3,2x_1+4x_2).
\]
As a minimum of linear functions, \(f\) is concave, and for \(p_1=p_2=1\), the optimum over the budget set \(x_1+x_2\leq b\) is attained at
\[
x_1=4b/31, x_2=1b/3,
\]
that is
\({dx_i}/{db}=1/3\).
 Is the optimum unique?

Consider the problem
\[
\begin{array}{ccc}
x&+y&\to\min, \mathrm{subject\ to}\\
x&+2y&\leq 2\\
x&&\geq 0\\
&y&\geq 0\\
\end{array}.
\]
Solve it.
Answer: \(x=2,y=0, p_*=2\).

Formulate and solve the dual problem.
Answer: the original problem is equivalent to
\[
\begin{array}{ccc}
x&+y&\to\min, \mathrm{subject\ to}\\
x&2y&\geq 2\\
x&&\geq 0\\
&y&\geq 0\\
\end{array},
\]
for which the dual is
\[
\begin{array}{cc}
2\lambda &\to\max, \mathrm{subject\ to}\\
\lambda &\leq 1\\
2\lambda &\leq 1\\
\lambda &\geq 0\\
\end{array}.
\]
Solution, is, obviously, \(\lambda=1, d_*=2\).

Solve it.
For problem 1, the Fenchel conjugate should be defined in terms of sup, right? If it is inf, there should be a negative sign.
Thank you! Corrected.
Does Problem 2 have 3 separate subsets? Or is that one subset defined with three inequalities? It seems that each inequality supersedes the previous one.
Defined by three inequalities: they are not implying each other.
Hi Professor,
Will you post solutions to the mock exam?
Answers posted.
Problem 4: could we assume x_i > 0 , \forany i?
Also, I think optimum exists iff \sum_i a_i \le 1. Is it correct?
Yes, all variables are positive.
Hi Professor, for question 2, I think the set may not be convex.
(0,0,4) and (2,2,0) both satisfy the three inequalities, so they are in the given set.
However, point (1,1,2), as a convex combination of these two points, does not satisfy the constraint a^3 – a(b^2+c^2) >= 0. Hence it is not in the given set. As it violates the definition of convex set, the given set is not convex.
Right – should have been strict inequalities.
Hi, professor.
I think the solution to Problem4(c) should be x1= (4b/3)1. If x1=(4b/3) and x2=1(b/3), x1 + x2 equals 1+b which is greater than b, which violates the restriction.
Thank you! – corrected.