Operator norm using spectral theorem

Let $\Omega=[0,1]$, consider the operator $A:f \rightarrow \int_0^x f(y) dy$ where $f \in L^2(\Omega)$. Compute $||A||$. Hint: Apply the spectral theorem to $A^* A$.

I have already shown that $A: L^2(\Omega) \rightarrow L^2(\Omega)$ and the adjoint operator $A^* f(x) = \int_x^1 f(y)dy$. The spectral theorem states that there is an orthonormal basis consisting of eigenvectors of a compact and self-adjoint operator. And I know that the largest eigenvalue equals the norm of the operator, but I don't know how to compute the eigenvalue and how to make sure the found eigenvalue is the largest. Could anyone help?

Here's an outline of the proof: Suppose $\alpha \neq 0$ is an eigenvalue of $A^*A$. Then there's an $f \in L^1[0,1]$ such that $A^* Af = \alpha f$. You can prove that $f$ is twice differentiable and obeys the differential equation $$\alpha f'' = - f.$$ Then we must have $f(x) =Ae^{i \omega x} + B e^{- i \omega x}$ for some $A, B$ and $\omega$ with $\omega^2=\frac{1}{\alpha}$. Then try to calculate $A^*Af$ and compare with $\alpha f$. It's a pretty tedious calculation so I won't do it here, but it will lead you to conclude that $A=B$. So you might as well write $f(x) = C \cos(\omega x)$. Lastly note that $A^*A(f)(1)=0$ so you must have $\cos(\omega) = 0$. This means $\omega$ can only be of the form $\omega = (n+1/2)\pi$ with $n \in \mathbb{Z}$. So we must have that the eigenvalue is equal to $\alpha = \frac{1}{\omega^2} = \frac{4}{\pi^2} \frac{1}{(1+2n)^2}$. The largest such eigenvalue is of course $\frac{4}{\pi^2}$.
Suppose $A^*Af=\lambda f$ for some $\lambda > 0$ and $f\ne 0$. Then $$\int_{x}^{1}\int_{0}^{y}f(z)dzdy = \lambda f.$$ Differentiating twice gives $$-f(x) = \lambda f''(x),\;\; f(1)=0, f'(0)=0.$$ $f$ satisfies the differential eigenvalue equation iff $$f(x)= A\sin((x-1)/\sqrt{\lambda}) \\ \mbox{ and } f'(0)=0.$$ The second condition gives the positive eigenvalues $$\lambda_n = 1/(n+1/2)^2\pi^2,\;\; n=1,2,3,\cdots.$$ The norm $\|A\|$ is the square root of the largest eigenvalue of $A^*A$, which is $2/\pi$.
• I guess we should first show that $f$ is $C^2$. And a small correction, $f(x) = a e^{\mathbf{i} \omega x } + b e^{- \mathbf{i} \omega x }$. And at the end we would get $\lambda_n = \frac{1}{(n+1/2)^2\pi^2}$. I would accept @Demophilus answer. But Thank you. – user424674 Nov 9 '17 at 21:13
• @Philo : To be precise, the range of $A$ contains $f\in L^2$ iff $f$ is an equivalence class in $L^2$ which contains an absolutely continuous function $\tilde{f}$ such that $\tilde{f}' \in L^2$ and $\tilde{f}(0)=0$. In any such equivalence class $f$, there is at most one such $\tilde{f}$. So, to answer your question: no, $f$ would not necessarily be in $C^2$. – DisintegratingByParts Nov 9 '17 at 22:35
• @Philo : Thanks for the correction on $\lambda_n$. I fixed that. – DisintegratingByParts Nov 9 '17 at 22:40