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Consider $A$ is a linear operator: $E \rightarrow E$ on Banach space $ E=C([0,1], \mathbb{R}) $. $(Af)(t) = \int_0^1{(t^2+s^2)f(s)ds}$
Prove that $A$ is bounded and find its norm. My solution: $||Af||=\sup \limits_{t \in [0,1]}|(Af)(t)|=\sup \limits_{t \in [0,1]} |\int_0^1{(t^2+s^2)f(s)ds}| \leq \sup \limits_{t \in [0,1]} \int_0^1{|(t^2+s^2)f(s)|ds} \leq \sup \limits_{t \in [0,1]} (t^2+1) ||f|| = 2||f|| $
Hence $A$ is bounded.
Is my proof correct? And how to find the norm?

Olha
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    It is correct but you could easily find a tighter upperbound, so tight that it would be the norm. – Anne Bauval Jan 04 '23 at 23:05
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    Be careful, in order to find the norm, you should be as tight as you can be when taking inequalties. Here, you can get $|Af| \leq \frac 43 |f|$. Then, to find the norm, find a function $f$ that satisfies the equality cases of your inequalities. – Zag Jan 04 '23 at 23:05

1 Answers1

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For boundedness, note that for any $f \in C([0,1])$, $$|Af|=|\int_{0}^{1}(t^2+s^2)f(s)ds|\leq \int_{0}^{1}(t^2+s^2)|f(s)|ds\leq ||f||_{\infty}(t^2+\frac{1}{3})\leq \dfrac{4}{3}||f||_{\infty}$$ Thus $||A||\leq\dfrac{4}{3}$.

Now pick $f=1$, then $$||A(1)||=sup_{t \in [0,1]}|\int_{0}^{1}(t^2+s^2)ds|=sup_{t \in [0,1]}(t^2+1/3)=\dfrac{4}{3}$$

Hence $||A||=\dfrac{4}{3}$.

Evil Witch
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  • Thank you. May I ask one more question? If I had $E=C([0,2], \mathbb{R})$ and $(Af)(t)=\int_0^2{|(t^2+s^2)f(s)|ds}$, I would also take $f=1$, right? – Olha Jan 05 '23 at 11:01
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    I think yes. However, your norm will change, since for any $f \in C([0,2])$, $$|Af|\leq ||f||{\infty}(2t^2+8/3)\leq 32/3||f||{\infty}$$ – Evil Witch Jan 05 '23 at 18:33