Eigenvalues of integral operator Let $f \in L^{p}[0,1]$ for $p \in (1,\infty)$ and consider the operator $Tf(u) = \int_{0}^{u} f(t) dt$. We want to show that $T$ has no eigenvalues. An eigenvalue $\lambda \in \mathbb{R}$ means that there is a non-trivial solution to the equation
$$
\lambda f(u) = \int_{0}^{u} f(t) dt
$$
for all $t \in [0,1]$.
Note that this is easy to do when $f$ is assumed to be continuous. We differentiate both sides to find the $f$ has to be something exponential and realize that $f(0) = 0$ (thus no solution). However, when $f$ is simply in $L^{p}$ but not continuous we cannot do this. My thought is to use the density of continuous functions in $L^{p}$ to do this, but I'm not able to formalize the argument.
My "argument" flows like this: If for fixed $\lambda \in \mathbb{R}$ has the property that
$$
\int_{0}^{u} f(t) dt - \lambda f(u) = 0
$$
for all $u \in [0,1]$, this implies that there is a continuous $g$ such that $\int_{0}^{u} g - \lambda g = 0$, which is a contradiction. However, I'm not sure how to prove the existence of such a $g$. Is it even true?
 A: Continuity of the operator: Note that
$$Tf(x)=\int_0^xf(t)dt=\int_0^1f(t)\cdot\chi_{[0,x]}(t)dt$$
so
$$\|Tf\|_p^p=\int_0^1\int_0^1|f(t)|^p\cdot\chi_{[0,x]}(t)dtdx=\int_0^1\bigg(|f(t)|^p\int_0^1\chi_{[t,1]}(x)dx\bigg)dt=$$
$$=\int_0^1|f(t)|^p\cdot(1-t)dt\leq\int_0^1|f(t)|^pdt=\|f\|_p^p$$
so $T$ is bounded and $\|T\|\leq1$. For the eigenvalues, let $\lambda\neq0$ be a scalar so that there exists a non-zero function $f\in L^p[0,1]$ such that  $Tf(x)=\lambda f(x)$, so we have that
$$f(x)=\frac{1}{\lambda}\int_0^xf(t)dt$$
Observe that
$$|f(x)-f(y)|=\frac{1}{|\lambda|}\bigg|\int_x^yf(t)dt\bigg|\leq\frac{1}{|\lambda|}\int_x^y|f(t)|dt=\frac{1}{|\lambda|}\int_0^1|f(t)|\chi_{[x,y]}(t)dt\leq$$
(using Holder's inequality)
$$\leq\frac{1}{|\lambda|}\|f\|_p\cdot\big(\int_0^1\chi_{[x,y]}(t)dt\big)^{1/q}=\frac{\|f\|_p}{|\lambda|}\cdot|x-y|^{1/q}$$
so $f$ is in $\text{Lip}(\frac{\|f\|_p}{|\lambda|},\frac{1}{q})$ and in particular it is continuous. But since $f(x)=\lambda^{-1}\int_0^xf(t)dt$, it is not only continuous, but differentiable too! Taking derivatives we see that $\lambda f'=f$. This has a standard solution, something with a constant multiple of the exponential, I cannot recall at the moment. I believe after plugging this standard solution in you will immediately get to the conclusion that $f=0$, a contradiction.
A: Denote $T : L^p [0,1] \to L^p [0,1]$ the integral operator
$$
(Tf)(u) = \int_0^u f(t)\, dt.
$$
Consider an operator $T^n$. Its norm can be evaluated as follows:
$$
\begin{align*}
||T^n f||_p &= \sqrt[p]{\int_0^1 dx \, \left|\int_0^x dt_n\dots\int_0^{t_2}f(t_1)\, dt_1\right|^p} \leq \sqrt[p]{\int_0^1 dx \, \left[\int_0^x dt_n\dots\int_0^{1}|f(t_1)|\, dt_1\right]^p}\\
&\leq \sqrt[p]{\int_0^1 dx \, \left[\int_0^x dt_n\dots\int_0^{t_3}||f||_p \, dt_2\right]^p} = ||f||_p \sqrt[p]{\int_0^1 dx \, \left[\int_0^x dt_n\dots\int_0^{t_3}\, dt_2\right]^p}\\
&= ||f||_p \sqrt[p]{\int_0^1 dx \, \left[\frac{x^{n-1}}{(n-1)!}\right]^p} \leq \frac{||f||_p}{(n-1)!} \sqrt[p]{\int_0^1 dx \, x^{n-1}} \leq \frac{||f||_p}{(n-1)!},
\end{align*}
$$
so $||T^n|| \leq \frac{1}{(n-1)!}$. By the way, I used Holder's inequality in the second inequality. By Gelfand's formula for spectral radius $r(T)$
$$
r(T) = \lim_{n \to \infty} \sqrt[n]{||T^n||} \leq \lim_{n \to \infty} \frac{1}{\sqrt[n]{(n-1)!}} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0.
$$
In particular, each eigenvalue of $T$ (if they exist in the first place) is zero.
Suppose we have
$$
\int_0^u f(t) \, dt = 0.
$$
for some $f \in L^p [0,1]$. Then for any interval $X \subset [0,1]$ we have $\int_X f(t) \, dt = 0$. Thus $\int_X f(t) \, dt = 0$ for any measurable $X \subset [0,1]$, i.e. $f = 0$ in $L^p [0,1]$.
