Proof related to commuting self adjoint operators Let $E$ be a complex Hilbert space and $(A_1,...,A_n) \in \mathcal{L}(E)^n$.

I want to show that if the operators $A_k$ are commuting and self adjoint, then
  $$\sup_{(\lambda_1,...,\lambda_n)\in B_n}\left\|\sum_{k=1}^n\lambda_kA_k\right\|=\left\|\sum_{k=1}^nA_k^2\right\|^{1/2}$$
  with $B_n$ is the open unit ball of $\mathbb{C}^n$.

My attempt: In general (even if $A_k$ are  not commuting and self adjoint), for $(\lambda_1,\cdots,\lambda_n)\in \mathbb{C}^n$, we have
$$\left\|\sum_{k=1}^n\lambda_kA_k\right\|\leq \left(\sum_{k=1}^n|\lambda_k|^2\right)^{1/2}\left\|\sum_{k=1}^nA_k^*A_k\right\|^{1/2},$$
and thus
$$\sup_{(\lambda_1,...,\lambda_n)\in B_n}\left\|\sum_{k=1}^n\lambda_kA_k\right\|\leq \left\|\sum_{k=1}^nA_k^2\right\|^{1/2}.$$
But I'm facing difficulties to show the renverse inequality.
 A: Suppose $z_1,..,z_n$ are complex numbers. What is
$\sup_{\lambda \in B_1}\left|\sum_k\lambda_k z_k\right|$? Cauchy Schwarz shows that it is smaller than $\sqrt{\sum_k |z_k^2|}$, on the other hand using $\lambda =\frac z{\|z\|}$ it is larger than that, so it is equal to $\sqrt{\sum_k|z_k|^2}$. Together with the spectral theorem this calculation gives the result.
Since the $A_k$ are all commuting and self-adjoint, we have a projection valued measure $dP(x)$ and functions $f_k:\Bbb R\to \Bbb R$ so that $A_k=\int f_k(x) dP(x)$. It follows that
$$\sum_k \lambda_k A_k = \int \sum_k\lambda_k f_k(x) dP(x).$$
The norm of this expression is the essential supremum of $\left|\sum_k \lambda_k f_k(x)\right|$. Well, I want to do a "suppose the supremum is at $x$" kind of argument and use the above result to get what we want. So let
$$\|f\|_x = \lim_{r\to0}\mathrm{ess\, sup}\ \left|f\lvert_{B_r(x)}\right|$$
the "maximal value at $x$". This semi-norm admits the above trick, ie $\sup_\lambda\left\|\sum_k\lambda_k f_k\right\|_x=\|\sum_kf_k^2\|_x^{1/2}$. Then
$$\left\|\sum_k \lambda_k A_k\right\|=\mathrm{ess\,sup}\,\left|\sum_k\lambda_k f_k\right|=\sup_{x\in\Bbb R}\left\|\sum_k \lambda_kf_k\right\|_x$$
When we take the supremum over $\lambda$ we recover something that must be greater than
$$\sup_{x\in\Bbb R}\left\|\sum_kf_k^2\right\|^{1/2}_x=\left\|\sum_k A_k^2\right\|^{1/2}.$$
