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.