Question

Question 8 (Chapters 1-8) [1 x 14 14 marks For the statements bellow, say if they are true or false. If true, give a short mathematical proof, if false, give a counterexample. (h) If f : Rn → R is convex and h : R → Rnxn s strictly convex and nondecreasing, then ho f is strictly convex (i) If f is strictly convex, then it is coercive. ) If f : Rn → R is such that the level set {x E Rn : f(x)-a) is convex for every α E R, then f is convex. (k) If f is convex and coercive, then it is strictly convex.

Answer 1

Solution: Convex: $f:R^n\to R$ is convex if domain of f,
dom(f)
is a convex set

and $f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y) \forall x,y\in $dom(f)$, 0\leq \lambda\leq 1$ .

$f:R^n\to R$ is strictly convex if domain of f,
dom(f)
is a convex set

and $f(\lambda x+(1-\lambda)y)< \lambda f(x)+(1-\lambda)f(y) \forall x,y\in $dom(f)$, 0< \lambda< 1,x\neq y$

(h) Let $f:R^n\to R$ is convex and
h : R→ Rn×n
is strictly convex and non decreasing.

Composition Rules:(i) $g(x)=h(f(x))$ is convex if

f is convex; h convex non-decreasing

(ii) $g(x)=h(f(x))$ is strictly convex if

f is convex; h strictly convex non-decreasing.

Since h is non-decreasing, we have
l >0
. Since h is convex, we have
h" > 0
.

Since f is convex, we have
f"> 0
.

Therefore,
g' (x) = [h(f(x)))' = h'(f)f'

and $g''(x)=[h(f(x))]''=[h'(f)f']'=h''(f)(f')^2+h'(f)f''\geq 0$

Since f is convex and h strictly convex and non-decreasing'

Therefore $g(x)=h(f(x))$ is strictly convex.

Thus the statement (h) is true.

(i) A continuous function f(x) that is defined on all of $R^n$ is coercive if

lim f(x) +o
.

That is for any constant M>0 there exists a constant
RM> 0
such that

whenever $||x||>R_M$.

Therefore if f is strictly convex, then it need not be coercive.

So statement is false.

(j) Let $f:R^n\to R$ is such that the level set

$C_{\alpha}=\left \{ x\in R^n:f(x)\leq \alpha \right \}$ for every $\alpha \in R$.

If $f:R^n\to R$ is convex then $C_{\alpha}=\left \{ x\in R^n:f(x)\leq \alpha \right \}$ for every $\alpha \in R$ is convex but converse is false that is if $f:R^n\to R$ is such that the level set

$C_{\alpha}=\left \{ x\in R^n:f(x)\leq \alpha \right \}$ is convex for every $\alpha \in R$ then $f:R^n\to R$ need not

be convex. Therefore statement is false.

Epigraph of $f:R^n\to R$:

$epi(f)=\left \{ (x,t)\in R^{n+1}:x\in~dom(f), f(x)\leq t \right \}$

f is convex if and only if epi(f) is a convex set.

(k) If f is convex and coercive, then it is strictly convex.

So statement is true.