Is the set of norming vectors weakly closed?

Question

Let $\mathcal{H}$ be a separable Hilbert space, let $p\ge 2$, and consider a bounded linear operator $ T\colon \mathcal{H}\to L^p(\mathbb R^d). $

Is the set $M=\{f\in \mathcal H\ :\ \|Tf\|_{L^p}= \|T\|\|f\|_{\mathcal H}\}$ closed in the weak topology of $\mathcal H$?

Here $\|T\|= \sup\{ \|Tf\|_{L^p}\ :\ \|f\|_{\mathcal{H}}=1\}$. Note that $M$ is always norm closed.

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Motivation. This idea comes from Sobolev embeddings.

As a simple example, let $T$ be the identity mapping $T\colon H^1(\mathbb R)\to L^\infty(\mathbb R)$, where $\|f\|_{H^1}^2=\|f\|_{L^2}^2+\|f'\|_{L^2}^2$. In this case, letting $f_{y}(x):=e^{-|x-y|}$, it turns out that $$ M=\{ af_{y}\ |\ a\in\mathbb R,\ y\in\mathbb R\}. $$ This set is weakly closed in $H^1(\mathbb R)$. Indeed, if the sequence $a_nf_{y_n}$ converges weakly, then it is bounded, and so $a_n$ must be bounded. If $y_n\in\mathbb R$ is also bounded, then we can extract a strongly convergent subsequence and we are done; if $y_n$ is not bounded, then $a_n f_{y_n}\rightharpoonup 0$, and $0\in M$.

The same reasoning applies to the embedding $H^1(\mathbb R^d) \subset L^p(\mathbb R^d)$, where $p=2d/(d-2)$. In this case, $M$ is the set of all translates and dilates of $f(x)=(1+|x|^2)^{-\frac{d}p}$.

Remark. If $p=2$, then $M$ is a closed subspace of $\mathcal H$, and so it is weakly closed. Indeed, since $\lVert Tf\rVert_{L^2}=\langle T^\ast Tf| f\rangle$, $M$ is exactly the eigenspace of $T^\ast T$ corresponding to its dominant eigenvalue. The fact that there is a dominant eigenvalue is a consequence of the boundedness of $T$, hence of $T^\ast T$.

Answer 1

I would like to share some of my thoughts on the problem. I think that the proposition is too hard to prove, and maybe it is even false, at this level of generality. However, there is an assumption on $T$ that makes the proposition true: it is the following Brezis-Lieb property (refers to Lemma 2.6 of Lieb, Sharp constants in the Hardy-Littlewood-Sobolev and related inequalities, Annals of Mathematics 1983).

Property. We say that $T$ satisfies the Brezis-Lieb property if, for all $g_n\in M$ such that $g_n\rightharpoonup g$, it holds that $$\tag{1}\|Tg_n\|_p^p =\|Tg\|_p^p+\|T(g_n-g)\|_p^p+o(1).$$

Proposition. Let $\Omega$ be a measure space, let $p>2$, and suppose that the bounded operator $T\colon \mathcal H\to L^p(\Omega)$ satisfies the Brezis-Lieb property. Then $M$ is weakly closed.

Proof. Let $g_n\in M$, $g_n\rightharpoonup g$, and denote $r_n:=g-g_n$. Since $g_n\rightharpoonup g$ we have $$\tag{2} \|g_n\|_{\mathcal H}^2 = \|g\|_{\mathcal H}^2 + \|r_n\|_{\mathcal H}^2 + \epsilon_n, $$ where $\epsilon_n\to 0$. Since $g_n\in M$, by the Brezis-Lieb property we have $$ \tag{3} \begin{split}\|T\|^p\|g_n\|_{\mathcal H}^p&=\|Tg_n\|_p^p= \|Tg\|_p^p + \|Tr_n\|_p^p + \eta_n \\ &\le \|T\|^p\|g\|_{\mathcal H}^p + \|T\|^p \|r_n\|_{\mathcal H}^p + \eta_n,\end{split} $$ where $\eta_n\to 0$. Now, since $p>2$, for all $a\ne 0, b\ne 0$ we have the strict inequality $a^p+b^p <(a^2+b^2)^{p/2}$. So, assuming that all sequences converge, as we may up to a subsequence as they are all bounded, we have $$ \|T\|^p\lim \|g_n\|_{\mathcal H}^p < (\|T\|^2\|g\|_{\mathcal H}^2 + \|T\|^2\lim\|r_n\|_{\mathcal H}^2)^\frac{p}{2}=\|T\|^p\lim \|g_n\|_{\mathcal H}^p,$$ where we used (2) in the last identity, provided that both $\|g\|_{\mathcal H}\ne 0 $ and $\lim\|r_n\|_{\mathcal H}\ne 0$. This is clearly a contradiction.

We conclude one of $\|g\|_{\mathcal H}$ and $\lim\|r_n\|_{\mathcal H}$ must vanish. If that's $\|g\|_{\mathcal H}$, that means $g_n\rightharpoonup 0$. If that's $\lim\|r_n\|_{\mathcal H}$, that means $g_n\to g$ strongly in $\mathcal H$. In both cases, $g\in M$. $\Box$

Remarks.

1. This is the standard subadditivity argument used to prove the existence of extremizers; see for example Lemma 2.7 in the aforementioned paper of Lieb.
2. The Brezis-Lieb property is a kind of compactness assumption on $T$. All compact operators trivially satisfy it, because $g_n\rightharpoonup g$ implies $Tg_n\to Tg$ strongly. Less trivially, all Sobolev embeddings satisfy the Brezis-Lieb property. Indeed, if $T\colon H^s\to L^p$ is the identity mapping, by the Rellich compactness theorem a sequence $g_n\stackrel{H^s}{\rightharpoonup} g$ converges $L^2$-strongly on balls, and so it converges pointwise almost everywhere; and now, if $g_n\to g$ almost everywhere, then (1) is satisfied by the standard Brezis-Lieb lemma. As far as I understand, this very idea has been used by Fanelli, Vega and Visciglia to study maximizers for inequalities in harmonic analysis: see 1 and 2.
3. If $g_n\rightharpoonup g$ then $Tg_n\rightharpoonup Tg$, which, however, is not enough to conclude the Brezis-Lieb property (1). This paper of Adimurthi and Tintarev gives sufficient conditions for the Brezis-Lieb property to be satisfied with weak convergence alone.