2
$\begingroup$

I'm having some trouble with the following exercise:

Let $E=\mathcal C([a,b])$ be the set of continuous maps $f:[a,b]\to \mathbb R$ with the supremum norm. Let $x\in E$ and define $\mathfrak{I}x\in E$ as $$(\mathfrak{I}x)(t)=\int_a^tx(s)ds$$

Prove that:

  1. $\mathfrak I\in L(E)$ (The set of linear continuous maps from $E$ to itself with the norm $\|f\|=\sup\limits_{\|x\|\leq1} \|f(x)\|$) and evaluate $\|\mathfrak I\|$.

  2. Show that $\mathfrak I$ has no eigenvalues.

I was able to prove (2) and in the first one, I managed to prove that $\mathfrak I \in L(E)$ and although I was able to find an upper bound to $\|\mathfrak I\|$, I wasn't able to evaluate its precise value.

How can this be done?

$\endgroup$
3

2 Answers 2

4
$\begingroup$

You can bound: $$\left|\int_a^t x\right|\le(t-a)\|x\|\le(b-a)\|x\|$$

So, the norm of your operator is less than or equal to $b-a$. But, as Geetha mentions, taking $x$ to be a nonzero constant function shows the norm is precisely $(b-a)$.

That’s because if an operator $T$ has $\|T\|\le N$ and $\|Tx\|=N\|x\|$ for some $x$, then $\|T\|=N$. This is a straightforward fact.

$\endgroup$
2
$\begingroup$

A more general fact holds. Assume a linear operator $T: C[a,b]\to C[a,b]$ is positive, i.e. for $f\ge 0$ we have $Tf\ge 0.$ Then $T$ is bounded and $\|T\|=\|T1\|_\infty.$ Indeed for any $f\in C[a,b]$ we have $\|f\|_\infty 1\pm f\ge 0.$ Hence $$\|f\|_\infty T1\pm Tf=T(\|f\|_\infty 1\pm f)\ge 0$$ Thus $$|Tf|\le \|f\|_\infty T1$$ which implies $$\|Tf\|_\infty \le \|T1\|_\infty \|f\|_\infty\ {\rm and }\ \|T\|\le \|T1\|_\infty$$ The converse inequality $\|T\|\ge \|T1\|_\infty $ holds obviously, which means the norm is attained at $f\equiv 1.$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .