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.

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 $T^\ast T$.


1 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, $g\in M$. $\Box$


  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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.