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Yes, this is true for $W_0^{1,p}(\Omega)$ for open and bounded $\Omega \subset \mathbb R^n$ and $1 < p < \infty$. If $\Omega$ has some regularity, it also works for $W^{1,p}(\Omega)$. The argument is …

Let $X$ be a Banach space and $X^*$ its topological dual space. Let us define the property (WS): For all sequences $(x_n^*) \subset X^*$ and all $x^* \in X^*$, we have $$x_n^* \rightharpoonup x^* \qu …

Your assumptions are not sufficient to consider this mapping property. Consider $V = \mathbb{R}$ and the mapping $f(x) = x^3$. It maps bounded sets to bounded sets, but its Nemytskii operator $y \map …

My answer is for $p \in (2,4)$, the other case should follow similar. Let $t \in (0,1)$ be given, such that $$\frac1p = \frac{1-t}{2}+\frac{t}{4}.$$ I will go to use the following consequence of Höld …

This is not true in the case of $H^{-1}(\Omega) = H_0^1(\Omega)^*$ (real spaces), where $\Omega \subset \mathbb R^d$ is bounded: In https://math.stackexchange.com/questions/336834/decomposition-of-fu …

Such an extension is not always possible, a counterexample can be found in Section 2 of https://arxiv.org/abs/1812.02419v3.

In optimization, you typically have a function $f \colon X \to \mathbb{R}$ which you are going to minimize. In order to apply first-order optimality conditions or first-order methods, you would like t …

As mentioned by @TaQ, the embedding $W(V,H) \hookrightarrow C([0,T];V)$ is, in general, not true. However, if your estimate would be true, you can extend the embedding operator from the dense subspace …

This answer is strongly inspired by example 3.11 in the book by Bauschke and Combettes. Let $H = \ell^2$ and consider a sequence $\{\alpha_n\} \in (1,\infty)$ with $\alpha_n \searrow 1$. Define \begi …

This functional is sequentially weakly lower semicontinuous under fairly mild assumptions on $f$. We need that $f$ is non-negative, continuous and bounded from above. Let $u_n \rightharpoonup u$ in $ …

I already asked this on M.SE, but get no answers. Is there a (simple) example of a finely open set (i.e. w.r.t. the fine topology in potential theory) $O$ in $\mathbb R^n$, $n \ge 2$, which is not op …

If you switch from your perspective of functions to their epigraph, I think that you end up with a "closure operator", see Wikipedia. I am not sure about other examples, but if $\mathcal{F}$ consists …

This is not true. Take a sequence of points $(x_n) \subset D$ which converge to a point on the boundary (or even worse: whose accumulation points are $\partial D$). Then, for every point $x_n$ we can …

This is not true. Take $\Omega = (-1,1)$ and functions $u_M$ like (I hope that I got the constants right) $$ u_M(x) =\begin{cases} -M^2 (|x|-1)(|x|-1+1/M) + M & \text{for } |x| > 1-1/(2M) \\ 1/4 + M & …

It is not possible to define a normal derivative for all $u \in H^1(\Omega)$ which depends continuously on $u$. The reason is that all $C_c^\infty(\Omega)$ is dense in $H^1_0(\Omega)$, but all $u \in …