This is not really an answer, more of an extended comment, but it was getting too long. Besides, if I am mistaken then it is easier to downvote or correct something posted as an answer. $\newcommand{\norm}[1]{\Vert#1\Vert} \newcommand{\Real}{{\bf R}} \newcommand{\ptp}{\hat{\otimes}}$

I think that in the original question, if one drops the symmetry requirement, then the answer is negative in general. Here's my reasoning.

Claim #1: if X is a Banach space then the bilinear maps $\Psi:X\times X \to \Real$ which are bounded by the bi-sublinear functional $\beta(x,y) = \norm{x}\ \norm{y}$ are in one-to-one correspondence with the linear functionals on the projective tensor product $X\ptp X$ that have norm $\leq 1$.

(This is somehow the defining property of the proj. t.p. -- or at least, it characterizes it up to isometric isomorphism.)

Claim #2: there exists a Banach space $X$ and a closed subspace $E$ such that the natural restriction map $(X\ptp X)^* \to (E\ptp E)^*$ is not surjective.

(I admit that I can't recall the details of an example, but I think taking $X$ to be $L^1$ and $E$ to be the span of the Rademachers would work.)

Now, take some $\psi\in(E\ptp E)^\ast$ which is not in the image of the restriction map, and which satisfies $\norm{\psi}\leq 1$. Let $\Psi$ be the corresponding bilinear functional on $E$. Clearly $\Psi(x,y) \leq \norm{x}\norm{y}=\beta(x,y)$. But any bilinear extension of $\Psi$ to a $\beta$-dominated functional on $X\times X$ would correspond to an element of $(X\ptp X)^*$ which extends $\psi$, and our choice of $E$, $X$ and $\psi$ ensures this can't happen.

Edit: Since there is a conflict between this and Kofi's answer, let me make the following observation. The argument given by Kofi would, if valid, apply equally well to the problem of extending bilinear functionals defined on $E\times Y$ to bilinear functionals on $X\times Y$, while still preserving domination by the given bi-sublinear functional. (Here $E$ is a closed subspace of $X$, and $Y$ is another Banach space.)

Translated into the language of tensor products as above, this would say that the natural restriction map $(X\ptp Y)^\ast \to (E \ptp Y)^\ast$ is surjective. But by the Hom-tensor duality for the proj tp, this is equivalent to saying that every bounded linear operator from $E$ to $Y^\ast$ extends to a bounded linear operator from $X\to Y^\ast$. In particular, taking $Y=\ell^1$, $X=Y^*=\ell^\infty$, this would say that every closed subspace $E$ of $\ell^\infty$ is topologically complemented, and that is very very far from true.

This is not really an answer, more of an extended comment, but it was getting too long. Besides, if I am mistaken then it is easier to downvote or correct something posted as an answer. $\newcommand{\norm}[1]{\Vert#1\Vert} \newcommand{\Real}{{\bf R}} \newcommand{\ptp}{\hat{\otimes}}$

I think that in the original question, if one drops the symmetry requirement, then the answer is negative in general. Here's my reasoning.

Claim #1: if X is a Banach space then the bilinear maps $\Psi:X\times X \to \Real$ which are bounded by the bi-sublinear functional $\beta(x,y) = \norm{x}\ \norm{y}$ are in one-to-one correspondence with the linear functionals on the projective tensor product $X\ptp X$ that have norm $\leq 1$.

(This is somehow the defining property of the proj. t.p. -- or at least, it characterizes it up to isometric isomorphism.)

Claim #2: there exists a Banach space $X$ and a closed subspace $E$ such that the natural restriction map $(X\ptp X)^* \to (E\ptp E)^*$ is not surjective.

(I admit that I can't recall the details of an example, but I think taking $X$ to be $L^1$ and $E$ to be the span of the Rademachers would work.)

Now, take some $\psi\in(E\ptp E)^\ast$ which is not in the image of the restriction map, and which satisfies $\norm{\psi}\leq 1$. Let $\Psi$ be the corresponding bilinear functional on $E$. Clearly $\Psi(x,y) \leq \norm{x}\norm{y}=\beta(x,y)$. But any bilinear extension of $\Psi$ to a $\beta$-dominated functional on $X\times X$ would correspond to an element of $(X\ptp X)^*$ which extends $\psi$, and our choice of $E$, $X$ and $\psi$ ensures this can't happen.

Edit: Since there is a conflict between this and Kofi's answer, let me make the following observation. The argument given by Kofi would, if valid, apply equally well to the problem of extending bilinear functionals defined on $E\times Y$ to bilinear functionals on $X\times Y$, while still preserving domination by the given bi-sublinear functional. (Here $E$ is a closed subspace of $X$, as above, and $Y$ is another Banach space.)

Take the sub-bilinear functional on $X\times Y$ to be $(x,y)\mapsto \norm{x}\norm{y}$. Then, translated into the language of tensor products as above, this would say that the natural restriction map $(X\ptp Y)^\ast \to (E \ptp Y)^\ast$ is surjective. But by the Hom-tensor duality for the proj tp, this is equivalent to saying that every bounded linear operator from $E$ to $Y^\ast$ extends to a bounded linear operator from $X\to Y^\ast$. In particular, taking $E=Y^*$ and $X=\ell^\infty(\hbox{suitable set})$, this would say that $Y^\ast$ is complemented in any copy of $\ell^\infty(\hbox{suitable set})$ that contains it as a closed subspace. This is not the case.

This is not really an answer, more of an extended comment, but it was getting too long. Besides, if I am mistaken then it is easier to downvote or correct something posted as an answer. $\newcommand{\norm}[1]{\Vert#1\Vert} \newcommand{\Real}{{\bf R}} \newcommand{\ptp}{\hat{\otimes}}$

I think that in the original question, if one drops the symmetry requirement, then the answer is negative in general. Here's my reasoning.

Claim #1: if X is a Banach space then the bilinear maps $\Psi:X\times X \to \Real$ which are bounded by the bi-sublinear functional $\beta(x,y) = \norm{x}\ \norm{y}$ are in one-to-one correspondence with the linear functionals on the projective tensor product $X\ptp X$ that have norm $\leq 1$.

(This is somehow the defining property of the proj. t.p. -- or at least, it characterizes it up to isometric isomorphism.)

Claim #2: there exists a Banach space $X$ and a closed subspace $E$ such that the natural restriction map $(X\ptp X)^* \to (E\ptp E)^*$ is not surjective.

(I admit that I can't recall the details of an example, but I think taking $X$ to be $L^1$ and $E$ to be the span of the Rademachers would work.)

Now, take some $\psi\in(E\ptp E)^\ast$ which is not in the image of the restriction map, and which satisfies $\norm{\psi}\leq 1$. Let $\Psi$ be the corresponding bilinear functional on $E$. Clearly $\Psi(x,y) \leq \norm{x}\norm{y}=\beta(x,y)$. But any bilinear extension of $\Psi$ to a $\beta$-dominated functional on $X\times X$ would correspond to an element of $(X\ptp X)^*$ which extends $\psi$, and our choice of $E$, $X$ and $\psi$ ensures this can't happen.

This is not really an answer, more of an extended comment, but it was getting too long. Besides, if I am mistaken then it is easier to downvote or correct something posted as an answer. $\newcommand{\norm}[1]{\Vert#1\Vert} \newcommand{\Real}{{\bf R}} \newcommand{\ptp}{\hat{\otimes}}$

I think that in the original question, if one drops the symmetry requirement, then the answer is negative in general. Here's my reasoning.

Claim #1: if X is a Banach space then the bilinear maps $\Psi:X\times X \to \Real$ which are bounded by the bi-sublinear functional $\beta(x,y) = \norm{x}\ \norm{y}$ are in one-to-one correspondence with the linear functionals on the projective tensor product $X\ptp X$ that have norm $\leq 1$.

(This is somehow the defining property of the proj. t.p. -- or at least, it characterizes it up to isometric isomorphism.)

Claim #2: there exists a Banach space $X$ and a closed subspace $E$ such that the natural restriction map $(X\ptp X)^* \to (E\ptp E)^*$ is not surjective.

(I admit that I can't recall the details of an example, but I think taking $X$ to be $L^1$ and $E$ to be the span of the Rademachers would work.)

Now, take some $\psi\in(E\ptp E)^\ast$ which is not in the image of the restriction map, and which satisfies $\norm{\psi}\leq 1$. Let $\Psi$ be the corresponding bilinear functional on $E$. Clearly $\Psi(x,y) \leq \norm{x}\norm{y}=\beta(x,y)$. But any bilinear extension of $\Psi$ to a $\beta$-dominated functional on $X\times X$ would correspond to an element of $(X\ptp X)^*$ which extends $\psi$, and our choice of $E$, $X$ and $\psi$ ensures this can't happen.

Edit: Since there is a conflict between this and Kofi's answer, let me make the following observation. The argument given by Kofi would, if valid, apply equally well to the problem of extending bilinear functionals defined on $E\times Y$ to bilinear functionals on $X\times Y$, while still preserving domination by the given bi-sublinear functional. (Here $E$ is a closed subspace of $X$, and $Y$ is another Banach space.)

Translated into the language of tensor products as above, this would say that the natural restriction map $(X\ptp Y)^\ast \to (E \ptp Y)^\ast$ is surjective. But by the Hom-tensor duality for the proj tp, this is equivalent to saying that every bounded linear operator from $E$ to $Y^\ast$ extends to a bounded linear operator from $X\to Y^\ast$. In particular, taking $Y=\ell^1$, $X=Y^*=\ell^\infty$, this would say that every closed subspace $E$ of $\ell^\infty$ is topologically complemented, and that is very very far from true.