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Suppose we have $X=L^2([0,1];\mathbb{R})$ and

\begin{equation} T:X\rightarrow X, \ Tf(x)=\int_0^1x^2yf(y)dy. \end{equation} Show that $T$ is compact and determine $||T||.$

I already have that $||T||\leq \frac{1}{\sqrt{15}} $ but I dont know how I can choose a function that approximate this value from above or if there exists a $L^2$-function under which the norm is equal to $ \frac{1}{\sqrt{15}}$. Furthermore for the compactness: Can one conclude that the range of $T$ is the space which is generated by the polynom $x^2$ because one can show that $\int_0^1yf(y)dy$ is finite by using Hölders inequality and therefore the integral is only a constant so we would have a finite rank operator?

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Since$$Tf(x)=\int_0^1x^2yf(y)\,\mathrm dy=x^2\int_0^1yf(y)\,\mathrm dy,$$the range of $T$ is $1$-dimensional (it is equal to $\langle x^2\rangle$) and therefore $T$ is compact (every continuous operator with finite-dimensional range is compact).

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