Limit of an operator and eigenvalues can you help me how to solve this exercice? Define in $L^2(-1,1)$ $$ T_k[f](x):= \int_{-1}^1 \frac{(-1)^k}{(2k+1)!}(xy)^{2k+1} f(y)\,\,dy $$
i) show that $||T_k||\to 0$;
ii) find eigenvalues of $T_0$.
I don't know how to solve the first point, while I tried to solve the second point solving a Fredholm equation:
$\int_{-1}^1xyf(y)\,dy=\lambda f(x)$.
So I have $xC=\lambda f(x)$, where $C=\int_{-1}^1yf(y)\,dy$, multiply $x$ in $L^2$ and I find $(x,x) C= \lambda C$. So $\lambda=2/3$ is correct? 
 A: Using a very crude estimate, and with Cauchy-Schwarz near the end, 
\begin{align}
\|T_kf\|^2
&=\int_{-1}^1\left|\int_{-1}^1\frac{(-1)^k}{(2k+1)!}\,x^{2k+1}y^{2k+1}\,f(y)\,dy\right|^2\,dx\\ \ \\
&\leq\int_{-1}^1\left(\int_{-1}^1\frac{1}{(2k+1)!}\,|x|^{2k+1}|y|^{2k+1}\,f(y)\,dy\right)^2\,dx\\ \ \\
&\leq\frac{1}{(2k+1)!}\int_{-1}^1\left(\int_{-1}^1\,|f(y)|\,dy\right)^2\,dx\\ \ \\
&=\frac{2}{(2k+1)!}\left(\int_{-1}^1\,|f(y)|\,dy\right)^2\\ \ \\
&\leq\frac{2}{(2k+1)!}\left(\int_{-1}^1\,|f(y)|^2\,dy\int_{-1}^11^2\,dy\right)^2\\ \ \\
&=\frac{8}{(2k+1)!}\,\|f\|^2\xrightarrow[k\to\infty]{}0.
\end{align}
Your argument for the eigenvalue looks fine to me. 
A: $T_k[f](x):= \displaystyle \int_{-1}^1 \frac{(-1)^k}{(2k+1)!}(xy)^{2k+1} f(y)\,\,dy$
$= \displaystyle \int_{-1}^1 \frac{(-1)^k}{(2k+1)!}x^{2k+1}y^{2k+1} f(y)\,\,dy = \frac{(-1)^k}{(2k+1)!} x^{2k+1} \int_{-1}^1 y^{2k+1} f(y)\,\,dy ; \tag 1$
it is clear that
$y^{2k + 1} \in L^2(-1, 1); \tag 2$
thus, using Cauchy-Schwarz and the fact that $\vert x^{2k + 1} \vert \le 1$ for $x \in (0, 1)$,
$\left \vert \displaystyle \dfrac{(-1)^k}{(2k+1)!} x^{2k+1} \int_{-1}^1 y^{2k+1} f(y)\,\,dy  \right \vert = \left \vert \dfrac{(-1)^k}{(2k+1)!} \right \vert \vert x^{2k + 1} \vert \left \vert \displaystyle \int_{-1}^1 y^{2k+1} f(y)\,\,dy  \right \vert$
$\le \left \vert \dfrac{(-1)^k}{(2k+1)!} \right \vert \vert x^{2k + 1} \vert \left \Vert y^{2k + 1} \right \Vert \left \Vert f(y) \right \Vert \le \left \vert \dfrac{(-1)^k}{(2k+1)!} \right \vert \left \Vert y^{2k + 1} \right \Vert \left \Vert f(y) \right \Vert, \tag 3$
where $\Vert \cdot \Vert = (\cdot, \cdot)^{1/2}$ deinotes the usual norm on $L^2(-1, 1)$; we combine (1) and (3):
$\Vert T_k[f](x) \Vert \le \left \vert \dfrac{(-1)^k}{(2k+1)!} \right \vert \left \Vert y^{2k + 1} \right \Vert \left \Vert f(y) \right \Vert; \tag 4$
we observe that 
$y \in (-1, 1) \Longrightarrow 0 \le (y^{2k + 1})^2 = y^{4k + 2}  \le 1, \tag 6$
and so
$\Vert y^{2k + 1} \Vert^2 = \displaystyle \int_{-1}^1 (y^{2k + 1})^2 \; dy \le \int_{-1}^1 1 \; dy = 2 \Longrightarrow \Vert y^{2k + 1} \Vert \le \sqrt 2; \tag 7$
in this light we see that (4) yields
$\Vert T_k[f](x) \Vert \le \left \vert \dfrac{(-1)^k}{(2k+1)!} \right \vert \sqrt 2 \left \Vert f(y) \right \Vert; \tag 8$
hence,
$\Vert T_k \Vert \le \left \vert \dfrac{(-1)^k}{(2k+1)!} \right \vert \sqrt 2 \to 0 \; \text{as} \; k \to \infty, \tag 9$
which solves part (i); as for part (ii), the eigen-equation for $T_0$ is
$T_0[f](x) =  x \displaystyle \int_{-1}^1 y f(y)\,\,dy = \lambda f(x), \tag{10}$
which shows that
$f(x) = \alpha x \tag{11}$
for some $\alpha \ne 0$; we must preclude $\alpha = 0$ since by definition eigenvectors cannot vanish; it then follows that
$\displaystyle \int_{-1}^1 y f(y) \; dy = \alpha \int_{-1}^1 y^2 \; dy \ne 0; \tag{12}$
now (10) also yields
$\left (\displaystyle \int_{-1}^1 y f(y)\,\,dy \right ) (x,  x) = \left (x,  x \displaystyle \int_{-1}^1 y f(y)\,\,dy \right ) = \lambda (x, f(x)) = \lambda \displaystyle \int_{-1}^1 yf(y) \; dy, \tag{13}$
and by virtue of (12) we conclude that
$\lambda = (x, x) = \displaystyle \int_{-1}^1 y^2 \; dy = \left ( \dfrac{y^3}{3} \right \vert_{-1}^1 = \dfrac{2}{3}, \tag{14}$
associated to the eigenvector given in (11).
