2-summing vs Hilbert-Schmidt norm for extended operator between Hilbert and Banach space

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.

• 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. – Dirk Werner Mar 15 at 19:23
• @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. – n_flanders Mar 16 at 4:54
• @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. – Dirk Werner Mar 16 at 11:29
• 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. – n_flanders Mar 16 at 12:01
• Thanks for your PS! – Dirk Werner Mar 16 at 17:55

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".