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I am looking for an example of a discontinuous embedding $H_1 \rightarrow H_2$, where $H_1,H_2$ are Hilbert spaces and $H_1 \subset H_2$. In other words, for any $C > 0$, we can find $x \in H_1$ such that $\Vert x \Vert _{H _1} = 1$ and $\Vert x \Vert _{H _2} > C$.

This wikipedia page gives an example of a discontinuous embedding $C^0([0,1], \mathbb{R}) \rightarrow C^0([0,1], \mathbb{R})$, the former equipped with the $L^1$ norm and the latter equipped with the $L^{\infty}$ norm. However, they are not Hilbert spaces. I am looking for an example for two Hilbert spaces.

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    $\begingroup$ What do you mean by "embedding"? Injective linear map? $\endgroup$ Sep 20, 2023 at 2:50

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In many "reasonable" situations, this is not possible due the closed graph theorem. Let us denote the embedding by $A$, i.e., $A \colon H_1 \to H_2$ is linear. Then, $A$ has a closed graph, provided that for all sequences $(x_k) \subset H_1$ with $$ x_k \to x \text{ in } H_1, \qquad A(x_k) \to y \text{ in } H_2,$$ we have $y = A(x)$. In many situations, this is satisfied and the closed graph theorem yields that $A$ is continuous. Note that this property just means, that if you have a sequence $(x_k) \subset H_1 \subset H_2$ which converges in $H_1$ and in $H_2$, then the limits have to coincide.

However, there are artificial examples, which do not satisfy the above, see here: https://mathoverflow.net/a/184471.

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  • $\begingroup$ Thanks. This is all I need to know. $\endgroup$
    – ryanstar
    Sep 20, 2023 at 13:44
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    $\begingroup$ This does not answer the question: 1) For a linear map $A:H_1\to H_2,$ bounded $\iff$ continuous $\iff$ closed graph, but this is off topic. The question is whether there exist an injective discontinuous $A.$ 2) Your link to mathoverflow does not provide a counterexample with Hilbert spaces. $\endgroup$ Sep 20, 2023 at 15:31
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    $\begingroup$ -1: the question asks for a discontinuous embedding and the answer takes as an assumption that the embedding is continuous. $\endgroup$ Sep 20, 2023 at 18:41
  • $\begingroup$ @AnneBauval: I do not understand your objections. 1) Yes, of course continuous is equivalent to closed graph. But my reasoning is that this fact is a serious obstruction to the existence of discontinuous embeddings, since most embeddings have a closed graph. 2) If you start with a Hilbert space $X$, the construction gives you an example with Hilbert spaces. $\endgroup$
    – gerw
    Sep 21, 2023 at 11:53
  • $\begingroup$ @MartinArgerami: Where did I assume a continuous embedding? I only explained that most embeddings are continuous because they have a closed graph. I also gave an example of a discontinuous embedding (or at least pointed to such a construction). $\endgroup$
    – gerw
    Sep 21, 2023 at 11:54

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