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?
Prove that $A$ is bounded and find its norm
functional-analysis
Best Answer
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 $||A||=\dfrac{4}{3}$.