$E = \ell^p$, with $1\le p<\infty$. For $x\in\ell^p$, $x = (x_1,x_2,\dots,x_n,\dots)$, check function
$$\varphi(x) = \begin{cases}\sum_{k=1}^{\infty}k|x_k|^2 &\text{ if } \sum_{k=1}^{\infty}k|x_k|^2<\infty \\
+\infty &\text{ otherwise}\end{cases}$$
is convex, l.s.c.
I have shown that $\varphi$ is convex, but I'm stuck in how to show it is l.s.c.. I attempt to use the definition to check that $\forall\lambda\in\mathbb{R}$, $A=\{\varphi\le\lambda\}$ is closed. Take a sequence $\{x^{(n)}\}$ in A s.t. $x^{(n)}\to x$ for some $x\in \ell^p$. I'm wondering how to use sum over power p to get the sum over squares. Or is there some other ways to prove it?
Best Answer
The function $\phi$ is the supremum of the functions $\phi_n(x) = \sum_{k=1}^n k |x_k|^2$. Each $\phi_n$ is lower semicontinuous, and a supremum of l.s.c. functions is l.s.c.