# Operator norm of integral operator

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

$$$$T:X\rightarrow X, \ Tf(x)=\int_0^1x^2yf(y)dy.$$$$ Show that $$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?

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).