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If $X$ is a topological space, then a functional $\varphi:X\to\mathbb{R}$ is lower-semicontinuous (l.s.c) if $\varphi^{-1}(a,\infty)$ is open in $X$ for any $a\in\mathbb{R}$. If $X$ is a Hilbert space, then $\varphi$ is weakly l.s.c if it is l.s.c on $X$ with its weak topology.

My question: If $X$ is a Hilbert space and $\varphi:X\to\mathbb{R}$, then $\varphi(x)\leq \liminf{x_n}$ whenever $x_n$ converges weakly to $x$ $\Rightarrow$ $\varphi$ weakly l.s.c?

I read this in "A invitation to Variational Methods in Differential Equations". It isn't a exercise, maybe a definition.

user127096
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Cristyan Pinheiro
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1 Answers1

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I think the author is being a little sloppy in this paragraph, which raised your question:

statement

Lower semicontinuity was earlier defined by the openness of upper level sets. The weak topology being non-metrizable, one should not casually insert "in other words" followed by the definition of sequential weak lower semicontinuity.

For example, Ciarlet (Mathematical Elasticity: Three-dimensional elasticity, Volume 1) is rather clear on this point.

Ciarlet

I have not seen anyone actually using weak lower semicontinuity. It's sequential weak lower semicontinuity that's being used all the time, and because of this, many authors drop "sequential" altogether. Unsurprisingly, confusion ensues.

Unfortunately I do not have an example of a sequentially weakly lsc functional on a Hilbert space that is not weakly lsc.

user127096
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