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](https://mathoverflow.net/questions/130519/absolutely-2-summable-operator-on-a-hilbert-space) 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.



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


