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Definition of the problem

Let $\mathcal{H}$ be a Hilbert space, $\dim\mathcal{H}\geq2$. Prove that the operator norm on $L\left(\mathcal{H}\right)$ is not induced by a scalar product.

We are hinted to prove that the orthogonal projection onto $span\left\{ \varphi\right\} $ for $\varphi\in\mathcal{H},\quad\left\Vert \varphi\right\Vert =1$, is given by $P_{\varphi}x=\left\langle x,\varphi\right\rangle \varphi,\quad x\in\mathcal{H}$. We have now to consider $P_{\varphi}$ and $P_{\psi}$, where $\varphi\perp\psi$ and $\left\Vert \varphi\right\Vert =\left\Vert \psi\right\Vert =1$.

My effort

I did prove the above citted statement.

My idea

I should use the parallelogram equality to show that the equality does not hold: $$ \left\Vert P_{\varphi}+P_{\psi}\right\Vert ^{2}+\left\Vert P_{\varphi}-P_{\psi}\right\Vert ^{2}=2\cdot\left\Vert P_{\varphi}\right\Vert ^{2}+2\cdot\left\Vert P_{\psi}\right\Vert ^{2}. $$

Further efforts

I did try to compute the first operator norm, and I get the following by using Pythagoreus Theorem and Cauchy-Schwarz, and since $\varphi\perp\psi$: $$ \begin{eqnarray*} \left\Vert P_{\varphi}+P_{\psi}\right\Vert ^{2}+\left\Vert P_{\varphi}-P_{\psi}\right\Vert ^{2} & = & \sup_{\left\Vert x\right\Vert =1}\left(\left\Vert P_{\varphi}x\right\Vert ^{2}+\left\Vert P_{\psi}x\right\Vert ^{2}+\left\Vert P_{\varphi}x\right\Vert ^{2}+\left\Vert -P_{\psi}x\right\Vert ^{2}\right)\\ & = & 2\sup_{\left\Vert x\right\Vert =1}\left(\left\Vert P_{\varphi}x\right\Vert ^{2}+\left\Vert P_{\psi}x\right\Vert ^{2}\right)\\ & = & 2\sup_{\left\Vert x\right\Vert =1}\left(\left\Vert \left\langle x,\varphi\right\rangle \varphi\right\Vert ^{2}+\left\Vert \left\langle x,\psi\right\rangle \psi\right\Vert ^{2}\right)\\ & \leq & 2\sup_{\left\Vert x\right\Vert =1}\left(\left|\left\langle x,\varphi\right\rangle \right|^{2}\left\Vert \varphi\right\Vert ^{2}+\left|\left\langle x,\psi\right\rangle \right|^{2}\left\Vert \psi\right\Vert ^{2}\right)\\ & = & 2\sup_{\left\Vert x\right\Vert =1}\left(\left|\left\langle x,\varphi\right\rangle \right|^{2}+\left|\left\langle x,\psi\right\rangle \right|^{2}\right)\\ & \leq & 2\sup_{\left\Vert x\right\Vert =1}\left(\left\Vert x\right\Vert ^{2}\left\Vert \varphi\right\Vert ^{2}+\left\Vert x\right\Vert ^{2}\left\Vert \psi\right\Vert ^{2}\right)\\ & = & 2\sup_{\left\Vert x\right\Vert =1}\left(\left\Vert x\right\Vert ^{2}+\left\Vert x\right\Vert ^{2}\right)\\ & = & 2\cdot\left(1+1\right)\\ & = & 4. \end{eqnarray*} $$

On the other side, by almost the same computations, we have: \begin{eqnarray*} 2\cdot\left\Vert P_{\varphi}\right\Vert ^{2}+2\cdot\left\Vert P_{\psi}\right\Vert ^{2} & = & \sup_{\left\Vert x\right\Vert =1}\left(2\left\Vert P_{\varphi}x\right\Vert ^{2}+2\left\Vert P_{\psi}x\right\Vert ^{2}\right)\\ & = & ...\\ & \leq & 4. \end{eqnarray*}

And I guess that would with difficulties prove my statement..

My Question

How would you go to solve the problem? Could you give me a few steps so I could go any further? Do you see any mistakes in what I did so far?

Thank you for your help, Franck.

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A closely related question: math.stackexchange.com/q/112844 –  Jonas Meyer Jun 18 '12 at 17:15

1 Answer 1

up vote 3 down vote accepted

You should get $\|P_\psi + P_\varphi\| = \|P_\psi - P_\varphi\| = 1$. Consider $x = \alpha \psi + \beta \varphi + y$ where $\alpha$ and $\beta$ are scalars and $y$ is orthogonal to $\psi$ and $\varphi$.

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Thank you Robert, your answer has been of a great help!! :) –  franckysnow Jun 18 '12 at 17:10

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