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The question is as follows:

Prove that the operator $A : C[0,1] \to C[0,1]$

$$(Ax)(t)= \int_{0}^{1} e^{t-s}x(s) ds$$ is indeed bounded and calculate its norm.

$\textbf{some attempt}:$

We have $||Ax||_{\infty} = \sup_{t\in [0,1]} \mid \int_{0}^{1} e^{t-s}x(s) ds \mid \leq \sup_{t\in [0,1]} \int_{0}^{1} \mid e^{t-s}x(s) \mid ds $

$\hspace{8.46cm} \leq \int_{0}^{1} e^{1-s} ds ||x(s)||_{\infty}$

Which will give us $||A||_{\infty} \leq (e-1)$, and is bounded.

Please let me know if I am wrong in my calculation?

And please let me know that how can I find its norm?

Thanks!

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You are right. Taking $x \equiv 1$, which has $||x||_\infty = 1$, gives $||A|| \ge |A(x)| = e-1$. Thus, $||A|| = e-1$.

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  • $\begingroup$ Many thanks! Now what will happen if we replace our operator with $(Ax)(t) = \int_{0}^{t} e^{-s}x(s) ds$, for $A :L_p[0,1] \to L_q[0,1]$? Can we use Holder inequality in this case? $\endgroup$ – Nikita Dec 8 '17 at 6:27

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