Let $m\in C[a, b]$. Consider on $(C[a, b], ||m||_∞)$ the Multiplication operator $A:C[a, b]\to C[a, b], Af=mf$. prove that $||A||=||m||_∞$.

Question

Let $m\in C[a,b]\;$.

Consider on $\big(C[a,b],\Vert m\Vert_{\infty}\big)$ the multiplication operator $\;A:C[a,b]\to C[a,b]\;,\;Af=mf\;$.

Prove that $\;\Vert A\Vert=\Vert m\Vert_{\infty}\;$.

My try:

Proof: $\Vert Af\Vert=\Vert mf\Vert\leqslant\Vert m\Vert\Vert f\Vert$

$\Vert A\Vert\leqslant\Vert m\Vert\quad\color{blue}{(1)}$

Take $f=1$.

Then $\;\Vert f\Vert=1\;$ and $\;Af=m\;.$

$\Vert Af\Vert=\Vert m\Vert\leqslant\Vert A\Vert\Vert f\Vert=\Vert A\Vert$

$\Vert A\Vert\geqslant\Vert m\Vert\quad\color{blue}{(2)}$

$(1)+(2)\implies\Vert A\Vert=\Vert m\Vert.$

Is this correct approach?

Comments would be appreciated.

Thanks.