Norm of an integral operator in $L^2$ Let T an operator in $L^2((-1,1))$ such that $Tf(x) = \int_{-1}^{1} (xy + y^2)f(y)dy$.
Calculate the norm of $T$. 
I tried calculating $\left\lVert Tf(x) \right\rVert$ first, but I got stucked with the square of the inside integral. Any help? 

Observe
\begin{align}
Tf(x) = x\int^1_{-1} dy\ yf(y)+\int^1_{-1}dy\ y^2f(y)=: \mu x +v
\end{align}
then we see that
\begin{align}
\|Tf\|_{2}^2 = \int^1_{-1} dx\ |\mu x+v|^2 =   \int^1_{-1}dx\ \mu^2x^2+\mu v x + v^2= \frac{2\mu^2}{3}+2v^2.
\end{align}
Next, observe that for any $f \in L^2(-1, 1)$, we can write
\begin{align}
f(x) = \sum^\infty_{n=0} c_n P_n(x)  = \frac{c_0}{\sqrt{2}}+ c_1\sqrt{\frac{3}{2}}x+c_2\sqrt{\frac{5}{8}}(3x^2-1)+\cdots
\end{align}
where $P_n$ is the normalized $n$th degree Legendre polynomial. Hence it follows
\begin{align}
\|f\|_2^2= \sum^\infty_{n=0}|c_n|^2
\end{align}
with 
\begin{align}
\mu =\int^1_{-1} dy\ yf(y) =\sqrt{\frac{2}{3}}c_1 
\end{align}
and
\begin{align}
v=\int^1_{-1} dy\ y^2f(y) = \frac{\sqrt{2}}{3}c_0+\frac{2}{3}\sqrt{\frac{2}{5}}c_2.
\end{align}
Finally, we see that
\begin{align}
\|Tf\|_2^2=&\ \frac{2}{3}\left(\sqrt{\frac{2}{3}}c_1 \right)^2+2\left(\frac{\sqrt{2}}{3}c_0+\frac{2}{3}\sqrt{\frac{2}{5}}c_2 \right)^2\\
\le&\ \frac{4}{9}c_1^2+2\left(\frac{2}{9}+\frac{8}{45}\right)(c_0^2+c_2^2)\\
\leq&\ 2\left(\frac{2}{9}+\frac{8}{45}\right)(c_0^2+c_1^2+c_2^2)= \frac{4}{5}(c_0^2+c_1^2+c_2^2)
\end{align}
which means
\begin{align}
\|T\|_\text{op} \leq \sqrt{\frac{4}{5}}\approx 0.894.
\end{align}
Consider $f(x) = P_0(x)+cP_2(x)$, then we see that
\begin{align}
\|Tf\|_2^2=2\left(\frac{\sqrt{2}}{3}+\frac{2}{3}\sqrt{\frac{2}{5}}c \right)^2=\frac{4}{5}(1+c^2) = \frac{4}{5}\|f\|_2^2
\end{align}
provided $c=\frac{2}{\sqrt{5}}$. Hence $\|T\|_\text{op} = \sqrt{4/5}$.
