# Examples of discontinuous embeddings between Hilbert spaces

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.

• What do you mean by "embedding"? Injective linear map? Sep 20, 2023 at 2:50

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.

• Thanks. This is all I need to know. Sep 20, 2023 at 13:44
• 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. Sep 20, 2023 at 15:31
• -1: the question asks for a discontinuous embedding and the answer takes as an assumption that the embedding is continuous. Sep 20, 2023 at 18:41
• @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.
– gerw
Sep 21, 2023 at 11:53
• @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).
– gerw
Sep 21, 2023 at 11:54