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Assume $(V,\| \|_V),(W,\| \|_W)$ are both finite dimensional normed spaces. We have the induced operator norm on $Hom(V,W)$.

When does it occur that this norm is actually induced from some inner product?

As observed by a comment of user225318, If $dimV=dimW=1$ then the answer can clearly be positive. It can be seen that in the case where $dimV=1 (V=\mathbb{R})$ and the norm on $W$ is induced by an inner product, the answer is positive. So let us require $dimV>1$ or that $\| \|_W$ is not induced by an inner product.

To state this more clearly, I would be happy to find a complete characterization, i.e necessary & sufficient conditions on the dimensions of $V,W$ and on their respective norms that are equivalent to the operator norm being induced by an inner product.

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    $\begingroup$ "Does it ever occur?" Of course yes. Let $V, W$ be $\mathbb{R}$ with the absolute value norm. Hom(V,W) is isomorphic to the real numbers with the absolute value norm, which can certainly be induced from some inner product. To make the question interesting you need to at least take the dimensions of V, W to be bigger than 1, and perhaps also specify that their norms are not induced via inner products. $\endgroup$
    – user225318
    Commented Apr 21, 2015 at 21:39
  • $\begingroup$ I agree about the dimensions. (I didn't think about the trivial case). I have edited the question accordingly. $\endgroup$
    – Asaf Shachar
    Commented Apr 21, 2015 at 21:45

1 Answer 1

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I assume here the spaces are finite dimensional (as in the question), and the vector spaces are real (for convenience, one can generalize to other fields too).

Recall first that a norm is induced from an inner product if and only if it satisfies the polarization identity: $$ 2\|u\|^2 + 2\|v\|^2 = \|u+v\|^2 + \|u-v\|^2 $$ for every $u,v\in V$.

Let $V, W$ be normed vector spaces.

Claim. If the norm on $W$ is not induced by an inner product, then the operator norm on $L(V,W)$ is not induced by an inner product.

Proof. Let $\phi:V \to \mathbb{R}$ be a linear functional such that $\sup_{v\in V\setminus \{0\}} |\phi(v)| / \|v\| = 1$. Take $w_1\neq w_2\in W$. The operators $w_j \phi$ are both in $L(V,W)$. It is easy to see that the operator norm $\|w_j \phi\|_{L(V,W)} = \|w_j\|_W$, and so the operator norm is induced by an inner product only if the norm on $W$ is induced by an inner product. Q.E.D.

In other words, one can isometrically embed $W \simeq L(\mathbb{R},W)$ into $L(V,W)$. So if the latter is an inner product space so is the former.

Claim. If the norm on $V$ is not induced by an inner product, then the operator norm on $L(V,W)$ is not induced by an inner product.

Proof. Fix $w_0\in W$. Then $L(V,\mathbb{R})$ embeds in $L(V,W)$ via $\phi \mapsto w \phi$. Hence if the norm on $L(V,W)$ is induced by an inner product, so must the operator norm on $L(V,\mathbb{R})$ which is the same as the dual space of $V$. This implies that $V'$ is an inner product space and hence so is $V$. Q.E.D.

In other words, one can isometrically embed $V' \simeq L(V,\mathbb{R})$ into $L(V,W)$, so if the latter is inner product, so is the former.

Claim. If $\dim V, \dim W > 1$ and $V,W$ are inner product spaces, the operator norm $L(V,W)$ is not induced by an inner product.

Proof. Let $v_1 \perp v_2 \in V$ and $w_1\perp w_2 \in W$ be unit vectors. Consider the mappings $u_1 = v \mapsto \langle v, v_1\rangle w_1$ and $u_2 = v \mapsto \langle v, v_2\rangle w_2$. One easily calculate that the operator norm $\|u_1\| = \|u_2\| = 1$. One calculates further that the operator norm $\|u_1 + u_2\| = \|u_1 - u_2\| = 1$. This violates the polarization identity. Q.E.D.

To conclude, $L(V,W)$ is an inner product space with respect to the operator norm iff both $V, W$ are inner product spaces and at least one of $V, W$ has dimension $1$.

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