Finding this operator's spectrum In an exam, my professor gave the following exercise:

State and prove the spectral theorem for compact operators. Let $K$ be the operator defined by:
$$Kf(t)=\int_0^1\min(t,s)f(s)\mathrm{d}s.$$
(i) Show $K$ is a bounded operator from $L^2([0,1])$ to itself;
(ii) Show $K$ is a compact and self-adjoint operator;
(iii) Determine the spectrum of $K$; is the point 0 an eigenvalue?
(iv) Calculate the operator norm of $K$.

Apart from wondering whether he meant compact self-adjoint operators or wanted statement and proof of the properties of the spectrum, i.e. 0 being an element and there being only a finite number of finite-multiplicity eigenvalues of $K$ outside any disk centered at 0, I can handle the proof part. This is a Hilbert-Schmidt operator, so we analyze the kernel. We show it is $L^2([0,1]^2)$ as follows:
\begin{align*}
\int_{[0,1]^2}\min(t,s)^2\mathrm{d}s\mathrm{d}t={}&\int_0^1\int_0^1\min(t,s)^2\chi_{t\leq s}(s)\mathrm{d}t\mathrm{d}s+{} \\
&{}+\int_0^1\int_0^1\min(t,s)^2\chi_{s\leq t}(s)\mathrm{d}s\mathrm{d}t={} \\
{}={}&\int_0^1\int_0^st^2\mathrm{d}t\mathrm{d}s+{} \\
&{}+\int_0^1\int_0^ts^2\mathrm{d}s\mathrm{d}t={} \\
{}={}&\int_0^1\frac{s^3}{3}\mathrm{d}s+\int_0^1\frac{t^3}{3}\mathrm{d}t={} \\
{}={}&\frac{1}{12}+\frac{1}{12}=\frac16.
\end{align*}
So the operator is both compact and bounded, and has operator norm at most $\frac16$. The kernel is evidently real and symmetric w.r.t. variable swapping, hence the operator is self-adjoint.
But how do I find the spectrum? How do I show 0 is(n't) an eigenvalue? And how can I find the spectral norm? I tried producing an eigenvector for zero, and should have proved monomials, $\sin(2\pi kx)$ and $\cos(2\pi kx)$ are not. So now I am a bit at a loss. I tried supposing $Kf=0$ and ended up with the following:
$$\int_0^tsf(s)\mathrm{d}s+t\int_t^1f(s)\mathrm{d}s=0,$$
for all $t\in[0,1]$. Now what? Then I tried using monomials to get the norm as being exactly the norm of the kernel. So I normalized $s^n$ in $L^2$, getting $\sqrt{2n+1}s^n$. I plugged that into $K$ and got $\frac{\sqrt{2n+1}}{n+1}t[1-\frac{t^{n+1}}{(n+1)(n+2)}$. Then I plugged this into wolfram, and got this, which evalued at $t=1$ gives this. Now the minimum appears to be 0, but then again it is attained for $n=-\frac12$ which gives $\frac{1}{\sqrt{x}}\notin L^2$, so at least this is good to go. But calculating that max gave me something around 0.3, which is more than $\frac16$ which I calculated for the norm of the kernel. So have I miscalculated something? And how do I find that spectrum?
 A: It is clear that $0$ is in the spectrum of $K$, since $K$ is compact. Let us address if it is an eigenvalue: if $$ Kf=0,$$ this means we have 
$$\tag{1}
0=t\int_t^1f(s)\,ds+\int_0^ts\,f(s)\,ds.
$$
Differentiating (via Lebesgue's Differentiation Theorem), 
$$\tag{2}
0=\int_t^1 f(s)\,ds-tf(t)+tf(t)=\int_t^1f(s)\,ds,\ \ \ \text{a.e.}.
$$ Then, for any $v,t\in[0,1]$,
$$
\int_t^vf(s)\,ds=\int_t^1f(s)\,ds-\int_v^1f(s)\,ds=0.
$$
Applying Lebesgue's differentiation again, we get that $f(t)=0$ a.e. It follows that $0$ is not an eigenvalue. 
For the nonzero part of the spectrum: since $K$ is compact, all nonzero elements of the spectrum are eigenvalues. Assume that $\lambda$ is an eigenvalue: then
$$\tag{3}
\lambda f(t)=t\int_t^1f(s)\,ds+\int_0^ts\,f(s)\,ds.
$$ 
The right-hand-side of $(3)$ is continuous, so $f$ is continuous. But, knowing that $f$ is continuous, the right-hand side is differentiable; so $f$ is differentiable. Repeating the reasoning, we get that $f$ is infinitely differentiable. If we differentiate $(3)$, we get
$$\tag{4}
\lambda f'(t)=\int_t^1 f(t)\,ds-tf(t)+tf(t)=\int_t^1 f(t)\,ds.
$$
Differentiating once again, 
$$\tag{5}
\lambda f''(t)=-f(t).
$$ The solution to this DE is 
$$\tag{6}
f(t)=\alpha\,\cos\frac{t}{\sqrt\lambda}+\beta\,\sin\frac{t}{\sqrt\lambda}.
$$
Putting this $f$ into $(3)$, we get, after simplifying,
$$
0=\alpha t\sqrt\lambda\sin\frac1{\sqrt\lambda}+\beta t\sqrt\lambda\cos\frac1{\sqrt\lambda}-\alpha\lambda.
$$
Evaluating at $t=0$, we get $\alpha=0$. So the equation will be satisfied when $\beta\ne0$ and for every $t>0;$ thus
$$
\cos\frac1{\sqrt\lambda}=0.
$$
This forces $$\frac1{\sqrt\lambda}=\frac\pi2+k\pi,$$ so the eigenvalues are 
$$
\lambda_k=\frac4{(2k+1)^2\pi^2},\ \ k\in\mathbb N.
$$

Finally, for the operator norm: since $K$ is selfadjoint, we have 
$$
\|K\|=\sup\{|\lambda|:\ \lambda\in\sigma(K)\}=\sup\left\{\frac4{(2k+1)^2\pi^2}:\ k\in\mathbb N\right\}=\frac4{9\pi^2}.
$$
