For an orthonormal basis ${(e_n)}_{n\geq1}$ of a Hilbert space $H$, and a bounded sequence ${(\beta_n)}_{n\geq 1}\subset\mathbb{R}$. How to show that the following equalities \begin{equation*} Te_n=\beta_ne_n,~n\geq1, \end{equation*} define a bounded linear operator $T:H\to H$ such that \begin{equation*} \|T\|=\sup_{n\geq1}|\beta_n|. \end{equation*}

I tried to prove it in one direction only: let $v\in H$ then $v=\sum\alpha_n e_n$ and $$\|T\|=\sup_{\|v\|=1}{\|Av\|}=\sup_{\|v\|=1}{\|T\sum\alpha_n e_n\|}=\sup_{\|v\|=1}{\|\sum\alpha_n Te_n\|}=\sup_{\|v\|=1}{\|\sum\alpha_n \beta_ne_n\|}$$ then $$\|T\|=\sup_{\|v\|=1}{\|\sum\alpha_n \beta_ne_n\|}\leq \sup_{n}\{|\beta_n|\}\sup_{\|v\|=1}{\|\sum\alpha_ne_n\|}= \sup_{n}\{|\beta_n|\}.1$$

but I can not prove the opposite direction


If $T$ is bounded, then for each $n$ we have $$|\beta_n|=\|\beta_ne_n\|=\|Te_n\|\leq\|T\|,$$ so the sequence $\{\beta_n\}$ is bounded, and we have $\sup_n|\beta_n|\leq\|T\|$.

  • $\begingroup$ But since the given basis is infinite dimensional ,how can we show that $T$ is linear? $\endgroup$
    – user429382
    May 15 '17 at 22:47
  • $\begingroup$ Let $x_i=\sum_n\alpha_{i,n}e_n$ ($i=1,2$) and let $\gamma$ be a scalar, and show that $T(x_1+\gamma x_2)=T(x_1)+\gamma T(x_2)$. $\endgroup$
    – Aweygan
    May 15 '17 at 23:08

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