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Let \begin{equation} p \in[1, \infty) \text { şi }\left(\alpha_n\right)_{n \in \mathbb{N}} \in l^{\infty} \end{equation} . Show that the linear operator \begin{equation} F: l^p \rightarrow l^p \end{equation} given by \begin{equation} F\left(\left(x_n\right)_{n \in \mathbb{N}}\right)=\left(\alpha_n x_n\right)_{n \in \mathbb{N}} \end{equation} for all \begin{equation} \left(x_n\right)_{n \in \mathbb{N}} \in l^p \end{equation} is continuous and it's norm is \begin{equation} \|F\|=\left\|\left(\alpha_n\right)_{n \in \mathbb{N}}\right\|_{\infty} . \end{equation}

I try to find a real number l such that the norm of F times x is less or equal to l times norm of x but i don't know how to prove this.I don't know if i'm on a good path.

Alexandru Harai
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  • A similar question in the more general setting of $L_p(X,\mathscr{B},\mu)$ spaces has been discussed here$\phi f \in L^p(\mu)$ whenever $f\in L^p(\mu)$, then $\phi \in L^\infty(\mu)$ – Mittens Oct 25 '22 at 18:41
  • If memory serves, the norm of $F$ is the supremum of the norm of $F(x)$ for $x$ of norm no greater than $1.$ This should get you where you want to go. – Chris Leary Oct 27 '22 at 14:25

1 Answers1

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Hint: Recall that a linear operator is continuous iff it is bounded and iff it is continuous at a single point.

stone327
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