Ytukyg

Let $X=L^{2}[0, 1]$ and $$Ax(t)=int_{0}^{1}{ts(1-ts)x(s)}ds.$$ I have shown that this operator is compact, hence all non-zero elements of the spectrum $sigma(A)$ are eigenvalues of $A$. But then I am stuck to find the eigenvalues in order to find the spectrum, i.e. all $lambda ne 0$ such that $$int_{0}^{1}{ts(1-ts)x(s)}ds=lambda x(t).$$ Can someone help?

Old, Naive, Too Complicated Solution.

Suppose that $x$ is an eigenfunction of $A$ w/ eigenvalue $lambda$. That is,
$$lambda x(t)=Ax(t)=int_0^1 ts(1-ts)x(s) ds.$$

See a remark in the comment section below that proves differentiability of $x$. Note that
$$lambda x'(t)=int_0^1 s(1-ts)x(s) ds-tint_0^1s^2x(s) ds.$$
That is,
$$lambda t x'(t)=lambda x(t)-t^2gamma(x)tag{1}$$
where $gamma:L^2[0,1]to Bbb C$ is given by $gamma(x)=int_0^1 s^2 x(s) ds$.

Observe that the eigenspace with the eigenvalue $0$ is given by
$$ker A=left{xin L^2[0,1]:int_0^1sx(s) ds=0wedge int_0^1s^2x(s) ds=0right}.$$
From now on suppose that $lambdane 0$.

From (1), we get
$$frac{d}{dt}frac{x(t)}{t}=-frac{1}{lambda}gamma(x).$$
Therefore,
$$frac{x(t)}{t}=C-frac{t}{lambda}gamma(x)$$
or
$$x(t)=Ct-frac{t^2}{lambda}gamma(x).$$
Since $gamma(x)=int_0^1s^2x(s) ds$, we need
$$gamma(x)=int_0^1s^2left(Cs-frac{s^2}{lambda}gamma(x)right) ds.$$
So
$$gamma(x)=frac{C}{4}-frac{1}{5lambda}gamma(x)implies C=left(4+frac4{5lambda}right)gamma(x).$$
Hence,
$$x(t)=gamma(x)Biggl(left(4+frac4{5lambda}right){t}-frac{t^2}{lambda}Biggr).tag{2}$$
Therefore, the eigenspace of $A$ w/ eigenvalue $lambda ne 0$ is a subspace of the span of
$$x_lambda(t)=left(4+frac4{5lambda}right){t}-frac{t^2}{lambda}.$$
Plugging this into $lambda x_lambda (t)=Ax_lambda (t)$, it turns out that $lambda$ must satisfy
$$lambdaleft(4+frac{4}{5lambda}right)=frac{frac{1}{lambda}+80}{60}.$$ That is, $lambda=frac{4pmsqrt{31}}{61}$.

Simpler Solution.

Suppose that $x$ is an eigenfunction of $A$ w/ eigenvalue $lambdaneq 0$. Then,
$$x(t)=frac{Ax(t)}{lambda}=left(frac{int_0^1 sx(s) ds}{lambda}right)t+left(-frac{int_0^1s^2x(s) ds}{lambda}right)t^2.$$
That is, $x(t)=at+bt^2$ for some constants $a,b$. Plugging this into $lambda x=Ax$, we get
$$lambda(at+bt^2)=left(frac{a}{3}+frac{b}{4}right) t +left(-frac{a}{4}-frac{b}{5}right)t^2.$$
Hence,
$$frac{a}{3}+frac{b}{4}=lambda awedge -frac{a}{4}-frac{b}{5}=lambda b.$$
So $lambda$ is an eigenvalue of the matrix
$$begin{pmatrix}frac13 & frac14\ -frac14 &-frac15end{pmatrix},$$ which means $lambda=frac{4pmsqrt{31}}{60}$.

Here is how I would approach this question: Keep differentiating both sides with respect to $t$ until you reach zero on one side. This will show that an eigenfunction has to be a polynomial of a certain degree. Work out the equations for the coefficients of that polynomial from the equation we got so far…

By the way: This operator seems to have a pretty large null space.