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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.

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