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Consider the space $V$ of continuous functions on [0,1] with the 2-norm $\|f\|_{2}^{2}=\int_{0}^{1}|f|^{2}$. Define a liner map $M_{\varphi} : V \rightarrow V$ by $M_{\varphi} f=\varphi f$ where $ \varphi \in C([0,1])$. Is $\mathcal{A}=\left\{M_{\varphi} | \varphi \in C([0,1])\right\}$ a Banach algebra?

Since $V$ is an incomplete normed linear space, it is not enough to prove that $\mathcal{A}$ is closed. I have proved that $M_{\varphi}$ is bounded and $\|M_{\varphi}\|=\|\varphi\|_{\infty}$. How to prove that $\mathcal{A}$ is a Banach algebra? Should I use the Stone–Weierstrass theorem?

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  • $\begingroup$ $\mathcal A$ is a subset of $C[0,1]$, which is a Banach space, so it is enough to show it is closed. $\endgroup$
    – GEdgar
    May 31, 2019 at 0:44
  • $\begingroup$ @freakish OP is looking at $C[0,1]$ with the $L^{2}$ norm, an incomplete space. He is looking at a family of operators on this space with the operator norm (I assume). $\endgroup$ Jun 4, 2019 at 8:27

1 Answer 1

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The operator norm of $M_{\phi}$ is the sup norm of $\phi$. Hence your space is isometrically isomorphic to $C[0,1]$ and this makes it a Banach space. Other properties of a Banach algebra are trivial verifications.

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