# 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.

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.