Is $\mathcal{A}=\left\{M_{\varphi} | \varphi \in C([0,1])\right\}$ a Banach algebra?

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?

• $\mathcal A$ is a subset of $C[0,1]$, which is a Banach space, so it is enough to show it is closed. May 31 '19 at 0:44
• @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). Jun 4 '19 at 8:27

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.