4
$\begingroup$

Let $(X, |\cdot|_1)$ and $(Y, |\cdot|_2)$ be Banach spaces. Let $\mathcal L(X, Y)$ be the space of all continuous linear maps from $X$ to $Y$. We endow $\mathcal L(X, Y)$ with the operator norm $\|\cdot\|$. Then $(\mathcal L(X, Y), \|\cdot\|)$ is a Banach space. I would like to ask if the following statement is true, i.e.,

Statement If $X, Y$ are Hilbert spaces, then so is $\mathcal L(X, Y)$.

If $Y =\mathbb R$, the statement is true by Riesz representation theorem. Thank you so much for your elaboration!

$\endgroup$
0

2 Answers 2

6
$\begingroup$

The parallelogram-identity fails already for $2\times 2$-matrices: Consider the Hilbert space $X=Y=\mathbb{R}^2$. Note that for symmetric matrices $\|A\|= r(A)$ ($r$ the spectral radius). Set $A=\rm{diag}(2,1)$ and $B=\rm{diag}(1,2)$. Then $$ \|A+B\|^2+\|A-B\|^2=3^2 + 1^2 =10, $$ but $$ 2(\|A\|^2+\|B\|^2)=2(2^2+2^2)=16. $$

$\endgroup$
3
$\begingroup$

Assume the operator norm is associated with an inner product. For a nontrivial projection $P$ we have $$1=\|(I-P)\pm P\|^2 \\ =\|I-P\|^2 +\|P\|^2\pm 2{\rm Re}\,\langle I-P,P\rangle\\ =2 \pm 2{\rm Re}\,\langle I-P,P\rangle $$ Hence $$2{\rm Re}\,\langle I-P,P\rangle=\pm 1$$ a contradiction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .