Compute the norm of the linear map $\phi:\mathcal C^0([0,1],\mathbb R) \to \mathcal C^1([0,1],\mathbb R)$

functional-analysislinear-transformations

I'm doing this exercise in a lecture about linear maps between Banach spaces.

Exercise 16.(src) Let $E=C([0,1],\mathbb{R})$ with $\|f\|_{\infty}$. Let $\phi$ defined as
$$\phi(f)(x) = \int_{0}^{x} tf(t) \, \mathrm{d}t. $$

  1. Show that $\phi$ is continuous.
  2. Compute the norm of $\phi$.

While I'm able to do (1), I unable to compute the norm of $\phi$, which is $\|\phi\|$.

My questions:

  1. Could you please verify if my proof on (1) looks fine or contains logical gaps/errors?

  2. Please shed me some light so compute $\|\phi\|$!

Thank you so much for your help!

My attempt:

  1. By Fundamental Theorem of Calculus, $(\phi(f))'(x) = xf(x)$ for all $x\in [0,1]$ and thus $\phi(f) \in \mathcal C^1([0,1],\mathbb R)$. It is easy to verify that $\phi:\mathcal C^0([0,1],\mathbb R) \to \mathcal C^1([0,1],\mathbb R)$ is a linear map.

We have $$\begin{aligned} \|\phi(f)\|_\infty &= \sup_{x \in [0,1]} \left| \int_0^x t f(t) \, \mathrm{d}t \right| &&\le \sup_{x \in [0,1]} \int_0^x |t f(t)| \, \mathrm{d}t \\ &\le \sup_{x \in [0,1]} \int_0^x |f(t)| \, \mathrm{d}t && \le \sup_{x \in [0,1]} \left (x \sup_{z \in [0,1]}|f(z)| \right) \\ &= \left (\sup_{z \in [0,1]}|f(z)|\right) \sup_{x \in [0,1]} \left (x \right) &&=\sup_{x \in [0,1]}|f(x)| \\ &= \|f\|_\infty\end{aligned}$$

As such, $$\|\phi\| = \sup_{f \in \mathcal C^0([0,1],\mathbb R)} \dfrac{\|\phi(f)\|_\infty}{\|f\|_\infty} \le \sup_{f \in \mathcal C^0([0,1],\mathbb R)} (1) =1$$

Since $\phi$ is a linear map and $\|\phi\| < \infty$, $\phi$ is continuous.

  1. We have $$\|\phi(f)\|_\infty = \sup_{x \in [0,1]} \left| \int_0^x t f(t) \, \mathrm{d}t \right| \quad \text{and} \quad \|f\|_\infty = \sup_{x \in [0,1]}|f(x)|$$

As such, $$\|\phi\| = \sup_{f \in \mathcal C^0([0,1],\mathbb R)} \dfrac{\|\phi(f)\|_\infty}{\|f\|_\infty} = \sup_{f \in \mathcal C^0([0,1],\mathbb R)} \dfrac{\sup_{x \in [0,1]} \left| \int_0^x t f(t) \, \mathrm{d}t \right|}{\sup_{x \in [0,1]}|f(x)|}$$

Best Answer

Let us assume that $\phi : E \to E$, where $E = C([0,1],\mathbb{R})$ as mentioned in the original problem. First, for each $x \in [0, 1]$ we get

$$ \left| \int_{0}^{x} t f(t) \, \mathrm{d}t \right| \leq \int_{0}^{x} t \| f \|_{\infty} \, \mathrm{d}t = \frac{x^2}{2}\|f\|_{\infty} \leq \frac{1}{2}\|f\|_{\infty}. $$

This shows that $\|\phi\| \leq \frac{1}{2}$. We claim that we actually have an equality here. To see this, note that

$$ \|\phi(1)\| = \sup_{x \in [0, 1]} \left| \int_{0}^{1} t \, \mathrm{d}t \right| = \sup_{x\in[0,1]} \frac{x^2}{2} = \frac{1}{2} = \frac{1}{2}\|1\|_{\infty}, $$

and so, $\|\phi\| \geq \frac{1}{2}$.

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