Is a continuous linear injective map between Hilbert spaces with dense image invertible?

Question

I'm trying to figure out whether the usual proof of Schur's lemma for linear representations can be generalized to unitary representations. To this end, I'm considering a continuous/bounded linear injective map $T:H\to H'$ which has dense image, where $H,H'$ denote complex Hilbert spaces.

(1) Is $T$ an isomorphism of Hilbert spaces? By isomorphism of Hilbert spaces I mean a linear map which preserves the inner product whose inverse is also linear and preserves the inner product. I'm aware that if $T$ is bijective, then $T^{-1}$ is linear and that if $T$ preserves the inner product, then $T^{-1}$ does as well. Hence, it suffices to show that $T$ is bijective and preserves the inner product. I have no clue how to prove/disprove this.

(2) If (1) is false: Is $T$ at least an isomorphism of topological vector spaces? I.e. does $T$ have a continuous inverse? I'm aware that by the bounded inverse theorem for Banach spaces, proving the bijectivity of $T$ would suffice to show this.

Answer 1

Let me give the following counterexample. For $H$ we choose the Sobolev space $H^1(\mathbb{R})$ (sometimes denoted by $W^{1,2}(\mathbb{R})$). This is the subspace of $L^2(\mathbb{R})$ which has weak derivatives that are again in $L^2(\mathbb{R})$. The norm is given by $$ \Vert f \Vert_{H^1(\mathbb{R})} = \sqrt{\Vert f \Vert_{L^2(\mathbb{R}}^2 + \Vert f' \Vert_{L^2(\mathbb{R})}^2},$$ respectively, the scalar product is given by $$ \langle f, g \rangle_{H^1(\mathbb{R})} = \int_\mathbb{R} \overline{f} g + \int_\mathbb{R} \overline{f'} g'. $$ Now consider the inclusion map $$ i: (H^1(\mathbb{R}), \Vert \cdot \Vert_{H^1(\mathbb{R})}) \rightarrow (L^2(\mathbb{R}), \Vert \cdot \Vert_{L^2(\mathbb{R})}, f \mapsto f. $$ The map is clearly injective and it is also bounded as $$ \Vert i(f) \Vert_{L^2(\mathbb{R})} = \Vert f \Vert_{L^2(\mathbb{R})} = \sqrt{\Vert f \Vert_{L^2(\mathbb{R})}} \leq \sqrt{\Vert f \Vert_{L^2(\mathbb{R}}^2 + \Vert f' \Vert_{L^2(\mathbb{R})}^2} = \Vert f \Vert_{H^1(\mathbb{R})}.$$ However, the map is not surjective. For this we need to exhibit an element $L^2(\mathbb{R}) \setminus H^1(\mathbb{R})$. By Morrey's inequality all functions in $H^1(\mathbb{R})$ admit a Hölder continuous representative, thus, $1_{[0;1]}$ would be an example (of course we could check this in a more elementary fashion, but it is good to know Morrey's inequality).

As mentioned in the comments above, the inclusion map is not even an isomorphism of normed spaces on its image. If it was, then there would exist a constant $C>0$ such that for all $f\in H^1(\mathbb{R})$ holds $$ \Vert f \Vert_{H^1(\mathbb{R})} \leq C \Vert f \Vert_{L^2(\mathbb{R})}. $$ For this start with your favourite function $f\in C_c^\infty(\mathbb{R})\setminus \{0\}$ (note that $C_c^\infty(\mathbb{R}) \subseteq H^1(\mathbb{R})$) and for any $\varepsilon>0$ define $f_\varepsilon(x) = \sqrt{\varepsilon} f(\varepsilon x)$. Change of variables tells us $$ \Vert f_\varepsilon \Vert_{L^2(\mathbb{R})} = \Vert f \Vert_{L^2(\mathbb{R})}. $$ However, the derivative picks up an additional factor of $\varepsilon$ and hence, we get $$ \Vert f_\varepsilon \Vert_{H^1(\mathbb{R})}^2 = \Vert f_\varepsilon \Vert_{L^2(\mathbb{R})}^2 + \Vert f_\varepsilon' \Vert_{L^2(\mathbb{R})}^2 = \Vert f \Vert_{L^2(\mathbb{R})}^2 + \varepsilon^2 \Vert f' \Vert_{L^2(\mathbb{R})}^2. $$ Thus, if the inverse of $i$ restricted to its image was continuous, we would have for any $\varepsilon>0$ $$ \Vert f \Vert_{L^2(\mathbb{R})}^2 + \varepsilon^2 \Vert f' \Vert_{L^2(\mathbb{R})}^2 \leq C^2 \Vert f \Vert_{L^2(\mathbb{R})}^2. $$ If $\Vert f' \Vert_{L^2(\mathbb{R})} \neq 0$, then we can send $\varepsilon \rightarrow \infty$ (well, bad naming... if somebody is offended by a large $\varepsilon$, feel free to edit) and get a contradiction. However, $f'$ is continuous, thus, if its $L^2$-norm vanishes, we get that $f'$ vanishes identically and hence, $f$ is constant. This is not possible as $f$ has compact support and is not identically zero.

Answer 2

(1) is not even true in the finite-dimensional case. For instance, consider $H=H'=\mathbb{C}$ and $T:\mathbb{C}\to\mathbb{C}$ given by $T(z)=2z$.

(2) is true in the finite-dimensional case, and more generally, as you mention, if you know that $T$ is surjective, then (2) is true by the open mapping theorem. However, $T$ does not have to be surjective. For a simple example, take $H=H'=\ell^2$ and let $T$ send a sequence $(x_n)$ to $(x_n/n)$. This is injective and bounded (indeed, $\|T(x)\|\leq\|x\|$). The image of $T$ is dense since it contains all sequences with finite support. But $T$ is not surjective: for instance, the sequence $(1/n)$ is not in its image, since the preimage would be the constant sequence with value $1$ but that sequence is not in $\ell^2$.