# Is the set of norming vectors weakly closed?

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 $$T^\ast T$$.

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.
• Jun 28, 2018 at 12:24