Let $\mathcal{L}(E)$ be the algebra of all bounded linear operators on a complex Hilbert space $E$.
On $\mathcal{L}(E)^2$, we have two equivalent norms:
\begin{eqnarray\*}
N\_1(A\_1,A\_2)
&=&\sup\left\{\|A\_1x\|^2+\|A\_2x\|^2,\;x\in E,\;\|x\|=1\;\right\},
\end{eqnarray\*}
and
$$N\_2(A\_1,A\_2)=\sup\left\{|\langle A\_1x,y\rangle|^2+|\langle A\_2x,y\rangle|^2,\;x,y\in E,\;\|x\|=\|y\|=1\;\right\}.$$



> 
> Assume that $A\_1A\_2=A\_2A\_1$ and $A\_1$ et $A\_2$ are normal operators on $E$. How to show that
> $$N\_1(A\_1,A\_2)= N\_2(A\_1,A\_2)?$$
> 
> 
> 


**My attempt:**


Notice that by the Cauchy–Schwarz inequality we have always $N\_2(A\_1,A\_2)\leq N\_1(A\_1,A\_2)$.


Now we aim to prove that the converse inequality holds when $A\_1A\_2=A\_2A\_1$ and $A\_1$ and $A\_2$ are normal operators on $E$. I tried to apply the spectral theorem.


Since $A\_1$ and $A\_2$ are commuting normal operators, il is well known that there exists a suitable measure space $(X,\mu)$; $\mu(X)<\infty$,
two functions $\varphi\_1,\varphi\_2\in L^\infty(\mu)$ and a unitary operator $U:E\longrightarrow L^2(\mu)$, such that each $A\_k$ is unitarily equivalent to multiplication by $\varphi\_k$, $k=1,2$. i.e.
$$UA\_kU^\*f=\varphi\_kf,\;\forall f\in E,\,k=1,2.$$
So, we can write
$$A\_kf=\varphi\_kf,\;\forall f\in L^2(\mu),\,k=1,2.$$
Hence,
$$\langle A\_kf\;,\;g\rangle=\langle \varphi\_kf\;,\;g\rangle=\int\_X\varphi\_k f\bar{g}d\mu,$$
and
$$\|A\_kf\|^2=\langle A\_kf\;,\;A\_kf\rangle=\langle \varphi\_kf\;,\;\varphi\_kf\rangle=\int\_X|\varphi\_k|^2|f|^2d\mu.$$


I am trying to solve the following question, but I did not reach to any answer, I would be so glad if anyone could help me on that.



Prove that $N_1(A_1,A_2)= N_2(A_1,A_2)$ when $A_1A_2=A_2A_1$ and $A_1$ and $A_2$ are normal operators
I will follow the OP's initial observation and notation. From the formula
$$\|A\_1 f\|^2+\|A\_2 f\|^2=\int\_X\left(|\varphi\_1|^2+|\varphi\_2|^2\right)|f|^2$$
it is clear that
$$N\_2(A\_1,A\_2)\leq N\_1(A\_1,A\_2)\leq\mathrm{ess}\sup\left(|\varphi\_1|^2+|\varphi\_2|^2\right).$$
Hence it suffices to show that, for any $\varepsilon>0$, we have
$$N\_2(A\_1,A\_2)>\mathrm{ess}\sup\left(|\varphi\_1|^2+|\varphi\_2|^2\right)-\varepsilon.$$
Let $C$ denote the essential supremum on the right-hand side. By definition, there is a set $U\subset X$ of positive measure on which $|\varphi\_1|^2+|\varphi\_2|^2$ is at least $C-\varepsilon/2$ (pointwise). This set has a subset $V\subset U$ of positive measure such that both $\varphi\_1(V)\subset\mathbb{C}$ and $\varphi\_2(V)\subset\mathbb{C}$ can be covered by a disk of radius $\sqrt{\varepsilon}/4$. Then,
$$\frac{1}{\mu(V)}\int\_V|\varphi\_k|^2-\left|\frac{1}{\mu(V)}\int\_V\varphi\_k\right|^2=\frac{1}{\mu(V)}\int\_V\left|\varphi\_k-\frac{1}{\mu(V)}\int\_V\varphi\_k\right|^2\leq\varepsilon/4$$
shows that
$$\left|\frac{1}{\mu(V)}\int\_V\varphi\_1\right|^2+\left|\frac{1}{\mu(V)}\int\_V\varphi\_2\right|^2\geq\frac{1}{\mu(V)}\int\_V\left(|\varphi\_1|^2+|\varphi\_2|^2\right)-\varepsilon/2\geq C-\varepsilon.$$
Choosing both $f$ and $g$ to be the $L^2$-normalized indicator function $\mathbf{1}\_V/\sqrt{\mu(V)}$, the previous inequality becomes
$$|\langle A\_1f,g\rangle|^2+|\langle A\_2f,g\rangle|^2\geq C-\varepsilon.$$
The proof is complete.


