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 they 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 $A_1$ et $A_2$ 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.

Thank you everyone !!!