Let $m\in C[a,b]$. Consider on $(C[a,b], \|\cdot \|_{\infty})$ the multiplication operator $A: C[a,b] \to C[a,b], \quad Af = mf$.

Prove that $\|A\| = \|m\|_{\infty}$.

In my book, we are given the following definition although it is written as an example:

A bounded sequence $\lambda = (\lambda_n)_{n \in \mathbb{N}}$ induces a multiplication operator $A_{\lambda} : \ell^2 \to \ell^2$ by $(A_{\lambda}f)(n):= \lambda_nf(n), \quad n \in \mathbb{N}, f \in \ell^2$.

Furthermore, the author provides a brief proof for elements of $\ell^2$ which I have modified to fit my assumptions which is shown below:

Since $\lambda$ is a bounded sequence, we have

$\|A_{\lambda}f\|_{\infty} = \sup\{\int_{a}^{b}|A_{\lambda}f(t)|dt\} \leq \sup\{\int_a^b \|A_{\lambda}\|_{\infty} |f(t)|dt\} = \|A_{\lambda}\|_{\infty} \|f\|_{\infty}$ for all $f \in C[a,b]$. So $A_{\lambda}$ is bounded and $\|A_{\lambda}\| \leq \|\lambda\|_{\infty}$.

On the other hand, $\|A_\lambda e_n\|_{\infty} = \|\lambda_n e_n\|_{\infty} = |\lambda_n| \|e_n\|_{\infty} = |\lambda_n|$ where $e_n$ is the $n$th standard unit vector. Hence, $\|A_{\lambda}\| \geq |\lambda_n|$ for every $n \in \mathbb{N}$, and thus $\|A_{\lambda}\| \geq \sup_n |\lambda_n| = \|\lambda\|_{\infty}$. Using both these estimates gives $\|A_m\| = \|m\|_{\infty}$.

Can somebody please explain to me if this is a proper imitation of the proof in $\ell^2$ for the supremum norm?

Thanks in advance, your assistance is greatly appreciated.

My textbook is Functional Analysis An Elementary Introduction by Haase.

  • $\begingroup$ I'm somewhat confused: if you were trying to prove $\|A_\lambda\| = \|\lambda\|_\infty$, why did you mention the original problem ($\|A\| = \|m\|_\infty$ for $A : C[a,b] \to C[a,b]$, $Af = mf$)? $\endgroup$ – kobe Feb 17 '15 at 17:54
  • $\begingroup$ I had seen a very similar solution involving the $\ell^2$ space and was trying to use that as a model for my problem. $\endgroup$ – Jamil_V Feb 17 '15 at 18:53

By the definition of operator norm, $$ \| A\|=\sup\{ \| Af\|_{\infty};\quad \| f\|_\infty \leq 1\}. $$ Note that $|m(t)f(t)|\leq \| m\|_\infty \| f\|_\infty$ $(\forall t\in [a,b])$. Hence $$ \| Af\|_\infty= \| m f\|_\infty\leq \| m\|_\infty \| f\|_\infty. $$ It follows that $$ \| A\|=\sup\{ \| Af\|_{\infty};\quad \| f\|_\infty \leq 1\}\leq \|m\|_\infty. $$ On the other hand, for the constant function $e(t)=1$ we have $\| e\|_\infty=1$ and $$ \| A e\|_\infty=\| m e\|_\infty=\| m\|_\infty. $$ Thus, $\| A\|=\| m\|_\infty.$

  • $\begingroup$ Thank you very much for clarifying, I see now exactly where my confusion was! $\endgroup$ – Jamil_V Feb 17 '15 at 20:15

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