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I am wondering if the following result holds true..

Suppose $T: L^2(\mathbb{R}) \mapsto L^2(\mathbb{R})$ is a continuous, injective linear operator, i.e. $T(\lambda f + g) = \lambda T(f) + T(g)$ where $f,g\in L^2(\mathbb{R})$ and $\lambda \in \mathbb{R}$, and $T(f) = T(g)$ implies that $f=g$ almost everywhere. Here, $L^2(\mathbb{R}) = \{f: \mathbb{R} \mapsto \mathbb{R} | \int_{\mathbb{R}} |f(x)|^2dx <\infty\}$.

Suppose I have a sequence $h_n$ and a function $h$ such that

$$ \|T(h_n) - T(h)\|_2 \rightarrow 0 $$

Then I know that along a subsequence $T(h_{n_k})$ that $T(h_{n_k}) \rightarrow T(h)$ almost everywhere.

Question: Does this also imply that $h_{n_k} \rightarrow h$ almost everywhere?

user61038
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  • If you know, in addition, that $T$ is bounded below (meaning that $|Tf|_2\ge C|f|_2$ for each $f\in L^2$), then at least the answer is yes. – Friedrich Philipp Oct 07 '17 at 02:42
  • hmm, that sounds like I would be assuming that the norm $|Tf|_2$ is equivalent to $|f|_2$ since that would imply by the continuity of $T$ that there is another constant $K$ such that $C|f|_2 \leq |Tf|_2 \leq K|f|_2$, where $K = |T|$, and $|T|$ is the operator norm. I'm wondering if there's a weaker condition that is sufficient enough to ensure convergence in that manner.. – user61038 Oct 07 '17 at 02:54
  • @FriedrichPhilipp: I have a counterexample where $T$ is an isometry - perhaps we are interpreting the question differently? – Dap Oct 07 '17 at 04:34
  • user61038: are you asking whether $h_{n_k}\to h$ a.e. for the same sequence (only assuming $T(h_{n_k})\to T(h)$ a.e.), or whether there exists a new sequence $n_k'$ with $h_{n_k'}\to h$ a.e? I think this ambiguity is why my answer contradicts @FriedrichPhilipp's comment. – Dap Oct 07 '17 at 14:40
  • I am asking whether the sequence $h_{n_k} \rightarrow h$ almost everywhere for the same sequence $T(h_{n_k}) \rightarrow T(h)$. I'm not sure I am seeing the contradiction, in your first example, you could take a subsequence that converges to 0 almost everywhere, I think. Representing the tuple $(n,m)$ as a single value $N$ (as you put it, we order the pairs $(n,m)$ in lexicographic order) then we could simply find a subsequence $N_k$ such that $h_{N_k}(x) = I_{[1/N_k, 2/N_k]}(x)$, in which case pointwise convergence is true. – user61038 Oct 08 '17 at 02:18

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Define indicator functions $$ h_{n,m}(x) = \mathbb{I}_{[m/n,(m+1)/n]}(x) $$ for integers $n\geq 1$ and for integers $m$ from $0$ to $n-1$. By ordering the pairs $(n,m)$ in lexicographic order we can think of $h_{n,m}$ as an ordinary sequence of functions and refer to limits as $(n,m)\to\infty$. Since $h_{n,m}$ converges to zero in $L^1$ and $L^2$, its Fourier transform converges to zero in $L^2$ and $L^\infty$. However, $h_{n,m}$ does not converge pointwise anywhere in $[0,1]$.

It is also possible to arrange that no subsequence of $h_n$ converges a.e. Take $h_n=\exp(2\pi n i x)\operatorname{rect}(x)$ where $\operatorname{rect}(x)=\mathbb{I}_{[-1/2,1/2]}(x)$. Every subsequence of $h_n$ diverges a.e.. Defining $T(f)(\xi)$ to be $\exp(-\xi^2)\hat f(\xi)$ where $\hat f$ is the Fourier transform, $T(h_n)$ converges to zero in $L^2$ and $L^\infty$.

Dap
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  • If $(f_n)$ converges to zero in $L^2$, then there exists a subsequence $(f_{n_k})$ that converges to zero a.e. – Friedrich Philipp Oct 07 '17 at 14:30
  • your second example has a function $h: \mathbb{R} \mapsto \mathbb{C}$. I'm looking at functions from $h:\mathbb{R} \mapsto \mathbb{R}$.. is there a similar example like this in this case? – user61038 Oct 08 '17 at 02:32
  • @user61038: sure, just take the real part (cos) – Dap Oct 08 '17 at 08:42