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Let $(e_n)$ be orthonormal basis of a Hilbert space $H$. Let $a=(\alpha_n)_{n=1}^\infty\subset \mathbb{K}$ be a sequence of scalars. I want to prove that an operator $P_a\colon H\to H$, defined as $$P_ax=\sum_{k=1}^\infty \alpha_k(x,e_k)e_k,\ \ \ x\in H$$ is with norm $\|P_a\|=\|a\|_{\infty}$, if $a\in \ell_\infty$. I was able to show that $\|P_ax\|\leq \|a\|_\infty \|x\|$ but I could use some help on finding the exact $x_0\in B_H$, such that $\|P_{a}x_0\|=\|a\|_\infty$.

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  • $\begingroup$ Assume the converse, that is $\|P_a\| < \|a\|_\infty$. There are two cases $|a_{n_0}| = \|a\|_\infty$ for some $n_0$ or there is a sequence $|a_{n_k}| \to \|a\|_\infty$. Can you obtain a contradiction? $\endgroup$ – user251257 Dec 2 '16 at 20:54
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Assume $\|P\| < \|a\|_{\infty}$. Choose a sequence $(k_{n})_{n\in\mathbb{N}}$ such that $|\alpha_{k_n}| \to \|a\|_{\infty}$ for $n \to \infty$. Define $x_n := e_{k_n}$, then you have $\|x_n\| = 1$ and $$\|P(x_n)\| = |\alpha_{k_n}| \to \|a\|_{\infty}, $$ a contradiction to $\|P\| < \|a\|_{\infty}$.

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