# Prove that the integral operator has no eigenvalues [closed]

Let $V$ be the vector space of all real valued continuous functions. Prove that the linear operator $\displaystyle\int_{0}^{x}f(t)dt$ has no eigenvalues.

• What are your thoughts? Have you tried anything? Jul 30, 2013 at 15:05
• Does this answer your question? $T$ has no eigen-values May 12, 2021 at 15:46

Suppose that $\lambda \in \mathbb{R}^*$ is an eigenvalue of this operator associated to the eigenvector $f \neq 0$. Then, for all $x$, you have :

$$\int_{0}^{x} f(t) \: dt = \lambda f(x)$$

with the condition that $f(0)=0$. If you differentiate the previous equality, you have $f(x) = \lambda f'(x)$ with $f(0) = 0$. You can easily solve the differential equation $f = \lambda f'$ but the only solution which satisfies $f(0)=0$ is the function $x \, \mapsto \, 0$.

• Why $f(0) = 0$ ? Oct 6, 2015 at 10:02
• @RisingStar : An eigenvector $f$ satisfies the relation : $$\forall x \in \mathbb{R}, \, \int_{0}^{x} f(t) \, dt = \lambda f(x)$$ with $\lambda \neq 0$. Evaluate this at $x=0$. Oct 6, 2015 at 10:27
• Actually why you have taken $\lambda$ to be non zero ? Oct 6, 2015 at 10:37
• @RisingStar : The case $\lambda = 0$ is quite straightforward. An eigenvector $f$ associated to $\lambda = 0$ would satisfy : $$\forall x \in \mathbb{R}, \, \int_{0}^{x} f(t) \, dt = 0$$ As a consequence, $f \equiv 0$, which is not possible since an eigenvector is not identically zero. Oct 6, 2015 at 11:31
• what do you mean by f is similar to 0? or you meant f=0? Nov 16, 2020 at 3:50

Hint: Suppose you have $\int_{0}^{x} f(t)dt = \lambda \cdot f(t)$. Differentiate both sides - you should easily be able to solve the resultant differential equation. Is your solution truly an eigenvector if it is nontrivial?

• You can't differentiate $f$ as it is assumed to be only continuous. May 9, 2016 at 21:23
• @Marra: there is no assumption that $f$ is differentiable. The LHS is differentiable, and so too must be $\lambda f(t)$. This is a consequence of $f$ being an eigenvector. May 9, 2016 at 21:30
• Indeed! But for that you must assume that $f$ is an eigenvector. thanks! May 9, 2016 at 21:32