Convexity and lower semi-continuity

Question

Let : $f : [0,1] \rightarrow (0,\infty)$ continuous function. For all $x \in [0,1]$ define

$$ \Lambda(x) = log \int_{0}^{1} e^{xy}f(y)dy,~~ \Lambda^{*}=sup_{z \in [0,1]}~(xz-\Lambda(z)).$$

(a)Prove that both $\Lambda $ and $\Lambda^{*}$ are convex function on $[0,1]$

(b)Show that $\Lambda^{*}$ is a lower semi-continuous function on $[0,1]$, i.e., for any sequence$(x_n)_{n\ge 1}$ in $[0,1]$ converging to $x\in[0,1]$ we have $lim\inf_{n\to\infty}\Lambda^{*}(x_n)\ge \Lambda^{*}(x)$

My Thought : I can solve the problem (a) by using the fact that $f$ is convex iff $f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}$.

When it comes to solve the problem (b), I already knew that convex function on the open interval is continuous, So I just have to show that the function is lower semi-continuous on the end points. But, I was not able to handle it.. Could you give me a few hints ??? and I want to know what the problem wants to say us. Thank you.!

Answer 1

Actually, the Fenchel Conjugate $f^*$ is always lower semi-continuous if $f$ is convex.

The epigraph of $\Lambda^*$ is \begin{equation} \operatorname{epi}(\Lambda^*) = \{(x,\mu) \mid \mu \ge xz - \Lambda(z), \forall z \in [0,1]\}. \end{equation} We can see that $\operatorname{epi}(\Lambda^*)$ is intersection of the collection of the following closed half-space: \begin{equation} (z, -1)^T (x, \mu) \leq \Lambda(z) \end{equation} Obviously, an arbitrary intersection of closed sets is closed. Thus $\operatorname{epi}(\Lambda^*)$ is closed, namely $\Lambda^*$ is closed.

Proof by contradiction, suppose $f$ is not l.s.c at $x_0$, i.e., there exists sequence $(x_k) \to x_0$ such that \begin{equation} \lim \inf_{k \to \infty} f(x_k) < f(x_0). \end{equation} In this sense, there exists a scalar $\gamma$ and a subsequence $\{x_k\}_\mathcal{K} \subset \{x_k\}$ such that \begin{equation} f(x_k) \leq \gamma < f(x_0), \forall k \in \mathcal{K}. \end{equation} since $f(x_k) = \sup_{z} zx_k - \Lambda(z) \leq \gamma$, thus $zx_k - \Lambda(z) \leq \gamma$ will be held for any $z \in [0,1]$. Then $\lim_{k \to \infty}zx_k - \Lambda(z) = zx_0 - \Lambda(z) \leq \gamma$ will be held for any $z$. Hence we have $\Lambda^*(x_0)=\sup zx_0 - \Lambda(z) \leq \gamma$, contradiction.