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It's written in my course of functional analysis that $L^p$ norms is lower-semicontinuous but not continuous.

For me, continuity of $\Phi: L^p\to \mathbb R$ is : if $f_n\to f$ in $L^p$ then $\Phi(f_n)\to \Phi(f)$. For $L^p$ space, it looks obviously continuous since $$|\|f_n\|_{L^p}-\|f\|_{L^p}|\leq \|f_n-f\|_{L^p}.$$ Therefore, if $f_n\to f$ in $L^p$, then obviously $$\|f_n\|_{L^p}\to \|f\|_{L^p}.$$

Is this wrong ?

user657324
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    Lower-semicontinous with respect to which norm/topology? Every norm is continuous with respect to itself, as your argument demonstrates when generalised. – Theo Bendit Apr 06 '19 at 07:25

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I guess you mean that it's semi lower continuous with respect to the weak topology, i.e. if $f_n\to f$ weakly, then indeed $$\liminf_{n\to \infty }\|f_n\|_{L^p}\geq \|f\|_{L^p},$$ and not better. The proof use the fact that since $x\mapsto |x|^p$ convex, then $$|x|^p\geq |y|^p+x^*(y)(x-y),$$ where $x^*(y)=p|y|^{p-2}y$ when $p\in (1,\infty )$ and $x^*(y)=\begin{cases}sgn(y)&y\neq 0\\ 1&y=0\end{cases}$, if $p=1$.

user659895
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