Can the induced matrix norms be induced by inner products?

Consider the induced matrix p-norm. My question is:

Can these norms be induced by inner product?

It sufficed to check that if it satisfies the parallelogram law. But I don’t find an immediate way to either find a counterexample or prove that it is true. I generated some examples using MATLAB. It seems that the answer should be no. Any idea?

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Theorem 1. Let $X$ be a normed space. For a given $z\in X$ and $f\in X^*$ define operator
$$
z\bigcirc f:X\to X:x\mapsto f(x) z
$$
Then we have the following equalities
$$(\lambda_1z_1+\lambda_2z_2)\bigcirc f=\lambda_1(z_1\bigcirc f)+\lambda_2(z_2\bigcirc f)$$
$$z\bigcirc (\mu_1 f_1+\mu_2 f_2)=\mu_1(z\bigcirc f_1)+\mu_2(z\bigcirc f_2)\tag{1}$$
$$\Vert z\bigcirc f\Vert=\Vert z\Vert\Vert f\Vert$$

Proof. Straightforward computation.

Theorem 2. Let $X$ be a normed space. Fix $f\in X^*$ with $\Vert f\Vert=1$, the map
$$
i:X\to\mathcal{B}(X):x\mapsto x\bigcirc f
$$
is an isometry.

Proof. Immediate consequence of theorem 1.

Theorem 3. If $X$ is a normed space with $\mathrm{dim}(X)>1$ then $\mathcal{B}(X)$ is not an inner product space.

Proof. If $\mathcal{B}(X)$ is an inner product space then does its every subspace. In particular $i(X)$ is a inner product space. Since $i$ is an isometry, then $X$ is an inner product space too. Now recall that we have isometric embedding
$$
j:X\to X^*:x\mapsto(z\mapsto\langle z, x\rangle)
$$
Since $\mathrm{dim}(X)>1$ we have two orthogonal vectors $e_1, e_2\in X$. It is not hard to check that for operators $A_1=e_1\bigcirc j(e_1)$, $A_2=e_2\bigcirc j(e_2)$ we have
$$
\Vert A_1+A_2\Vert^2+\Vert A_1-A_2\Vert^2\neq 2\Vert A_1\Vert^2+2\Vert A_2\Vert^2
$$
so $\mathcal{B}(X)$ is not an inner product space. Contradiction.