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Let $H$ be a Hilbert space and $b$ a continuous and symmetric bilinear form on $H \times H$, such that the induced operator $T \colon H \to H^{\ast}$ (the star denoting the continuous dual) is Hilbert-Schmidt. Suppose that $H$ is continuously and densely embedded in a Banach space $B$ and that

$$ \sup_{h_1,h_2 \in H \setminus \left\{ 0 \right\}} \frac{ \left| b(h_1,h_2) \right| }{ \left\lVert h_1 \right\rVert_H \left\lVert h_2 \right\rVert_B }\leq C $$

where the norms on the right are those associated to $H$ and $B$. Then $T$ can be extended to a continuous linear operator $T_1 \colon B \to H^{\ast}$. Also, if I'm not mistaken, the above inequality allows to consider $T$ as a continuous linear operator $T_2 \colon H \to B^{\ast}$.

My question: Can the 2-summing norm $\pi_2(T_2)$ of $T_2$ be estimated by the Hilbert-Schmidt norm $\left\lVert T \right\rVert_2$ of $T$, i.e. does there exist some (ideally explicit) $\varphi$ such that $\pi_2(T_2) \leq \varphi(|\left\lVert T \right\rVert_2)$? In the answer to this question, it was mentioned that

$$ \pi_2(T_2) = \sup \left( \sum_n^{} \left\lVert T_2e_n \right\rVert_{B^{\ast}}^2 \right)^{1/2}, $$

where the supremum is taken over all orthonormal bases $\left\{ e_n \right\}$ of $H$, which might be relevant here.

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  • $\begingroup$ I think what you are asking is this: If $H$ is a Hilbert space, $X$ ($=B^*$) is a Banach space, $T:H\to X$ is an operator, $i: X\to H$ is a continuous injection, can one estimate the 2-summing norm $\pi_2(T)$ by a multiple of $\pi_2(iT)$? Take $H=\ell_2$, $X=\ell_1$, $i=$ the identity mapping and $T=T_n=$ the projection onto the first $n$ coordinates. Then $\pi_2(iT_n)= \sqrt{n}$, but $\pi_2(T_n)$ is of order $n$; see 1.6.8 in Pietsch's Eigenvalues and $s$-Numbers. $\endgroup$ – Dirk Werner Mar 15 at 19:23
  • $\begingroup$ @DirkWerner Thanks for the comment, in general what I'm asking for can of course not hold, I was hoping that the special additional structure of $T_2$ would make a difference. Also, I should add that a non-linear dependence would be fine as well and have adjusted my question. In fact, in the very useful reference you're quoting there is Proposition 1.5.3, which says that if $T$ is 2-summing, $S$ is continuous and factors through a Hilbert space, then $T$ and $ST$ have the same 2-summing norm, which at least shows that $T_2$ has finite 2-summing norm. $\endgroup$ – n_flanders Mar 16 at 4:54
  • $\begingroup$ @nflanders I think my example, when normalised, fits your setup, with $b(x,y)= n^{-1/2}\sum_{k=1}^n x_k y_k$ so that $|b(x,y)|\le \|x\|_2\|y\|_\infty$ (i.e., $C=1$). So one should replace $T=T_n$ above with $T_n/\sqrt{n}$. -- I think you have misread Prop.\ 1.5.3 that does not say what you quote, but that $\pi_2(T)$ equals the (quasi-) norm of $T$ in the product operator ideal $\mathfrak{H}\circ \mathfrak{P}_2$, cf.\ ibidem D.1.10. $\endgroup$ – Dirk Werner Mar 16 at 11:29
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    $\begingroup$ PS: In case you happen to be the author of the famous German FA book, thanks for writing such a great text! I'm learning from it self-dependently and like it a lot. It is rigorous while also appealing at intuition, exactly as it should be. $\endgroup$ – n_flanders Mar 16 at 12:01
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    $\begingroup$ Thanks for your PS! $\endgroup$ – Dirk Werner Mar 16 at 17:55
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The question can be reformulates as follows. If $H$ is a Hilbert space, $X$ ($=B^∗$) is a Banach space, $T:H\to X$ is an operator, $i:X\to H$ is a continuous injection, can one estimate the $2$-summing norm $\pi_2(T)$ by a multiple of $π_2(iT)$?

The following example shows that this is in general impossible. Take $H=\ell_2$, $X=\ell_1$, $i=$ the identity mapping and $S=S_n=$ the projection onto the first $n$ coordinates and $T_n=S_n/\sqrt{n}$. This matches the OP's setup with with $b(x,y)=n^{−1/2} \sum_{k=1}^n x_ky_k$ so that $|b(x,y)| \le \|x\|_2 \|y\|_\infty$ (i.e., $B=c_0$ and $C=1$).

Then $\pi_2(iT_n)=1$, but $\pi_2(T_n)$ is of order $\sqrt n$; see 1.6.8 in Pietsch's "Eigenvalues and s-Numbers".

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