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Let T be an operator $$Tf(t) = \int_0^1K(t,s)f(s)ds$$ where $f \in C[0,1]$ and $K \in C([0,1]\times[0,1])$. Prove $T \in \mathcal{L}(C[0,1])$ and calculate its norm.

I think I have successfully proven that it is indeed bounded:

Taking arbitrary $f \in C[0,1]$ we have: $$ |Tf(t)| = \left|\int_0^1K(t,s)f(s)ds\right| \leq \int_0^1\left|K(t,s)f(s)\right|ds \leq $$ K is continous and on compact set so there is $M$ such that $|K| \leq M$ $$ \leq M \int_0^1\left|f(s)\right|ds \leq M(1-0)\|f\|_{\infty} $$ Now, taking supremum $\|f\|_{\infty} = 1$ we have: $$ \sup_{\|f\|_{\infty}=1}\|Tf(t)\|_\mathbb{\infty} \leq M \implies \|T\| \leq M < \infty $$ So $T$ is indeed bounded.

How do I calculate its norm?

blahblah
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  • Are you sure that the application $T$ goes from $C([0,1])$ to $\mathbb{R}$? It isn't $T:C([0,1]) \rightarrow C([0,1])$? – Falcon Nov 08 '20 at 11:17
  • Otherwise $T$ is not well-defined because it takes different values for different $t$. – Falcon Nov 08 '20 at 11:19
  • Um, the exercise says to prove that $T \in \mathcal{L}(C[0,1])$ and (if I understand correctly) $\mathcal{L}(C[0,1])$ means $\mathcal{L}(C[0,1], \mathbb{R})$ which is "set of linear bounded operators from $C[0,1]$ to $\mathbb{R}$. – blahblah Nov 08 '20 at 11:26
  • $\mathcal{L}(A)$ is the set of all linear bounded function from $A$ to $A$. Maybe edit your question. – Falcon Nov 08 '20 at 11:31
  • Done - I have edited the proof which still seems fine. – blahblah Nov 08 '20 at 11:47
  • Maybe this will help you https://math.stackexchange.com/questions/3711801/finding-norm-of-tft-int-01ks-tfsds-if-k-is-continuous-functional-def/3711949#3711949 – thing Nov 08 '20 at 16:07

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