Let $\mathcal{H}_1$ and $\mathcal{H_2}$ two Hilbert spaces. Construct the tensor product $\mathcal{H}_1 \otimes \mathcal{H}_2$ as the set of all bounded antilinear operators from $\mathcal{H_2}$ to $\mathcal{H_1}$ such that the norm $\Vert A \Vert_\otimes := \sqrt{\sum_\beta \Vert Av_\beta \Vert^2}$ is finite, where $\lbrace v_\beta \rbrace$ is any orthonormal basis of $\mathcal{H}_2$ (I followed the construction presented in Folland's A Course in Astract Harmonic Analysis).

Now for all $(u,v) \in \mathcal{H}_1 \times \mathcal{H}_2$ I can consider the antilinear operator $u \otimes v : w \mapsto \langle v, w \rangle u$, which is an element of the tensor product.

My question is: if I consider the map $\bigotimes : \mathcal{H}_1 \times \mathcal{H}_2 \rightarrow \mathcal{H}_1 \otimes \mathcal{H}_2$ defined in the natural way, is it continuous (w.r.t. the natural topologies on both spaces, i.e. the product topology on the domain and the topology induced by the norm on the codomain)? My guess is that it is continuous, but I can't prove it.

Denoting by:

  • $\pi_1$ and $\pi_2$ the two projections
  • $f_v = \langle v,\cdot \rangle \in \mathcal{H}_2^*$
  • $f : v \mapsto f_v$ the map which characterizes the dual space of $\mathcal{H}_2$

I tried to say that $\bigotimes(\cdot) = f_{\pi_2(\cdot)}\pi_1(\cdot)$ but I'm confused and can't tell if this is sufficient to conclude that $\bigotimes$ is continuous.

  • $\begingroup$ You are missing squares on both sides of the definition of $\|\cdot\|_\otimes$. Furthermore, $H_1\otimes H_2$ is not the set of all bounded antilinear operators, but only those for which $\|\cdot\|_\otimes$ is finite. $\endgroup$ – MaoWao Jan 22 '19 at 15:05
  • $\begingroup$ Thanks, I fixed it! $\endgroup$ – M. Rinetti Jan 22 '19 at 15:18

The map $\otimes$ is obviously bilinear. A bilinear map $\beta\colon H_1\times H_2\to K$ is continuous if and only if there exists a constant $C>0$ such that $\|\beta(u,v)\|\leq C\|u\|\|v\|$ for all $u\in H_1$, $v\in H_2$. The proof is very similar to the characterization of continuous linear maps as bounded operators.

In your case one has $\|u\otimes v\|_\otimes=\|\langle v,\frac{v}{\|v\|}v\rangle u\|=\|u\|\|v\|$, which shows that $\otimes$ is continuous.

  • $\begingroup$ Thank you, it is all clear now! $\endgroup$ – M. Rinetti Jan 22 '19 at 15:27

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