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 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.
 A: 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.
