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.