Integral operator has no eigenvalue

Question

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. This question is same as Prove that the integral operator has no eigenvalues However i am not able to understand why $f(0)=0$

Answer 1

Suppose that $$ \displaystyle\int_{0}^{x}f(t)dt=\lambda f(x), $$ then we can differentiate both sides: $$ f(x)=\lambda f'(x)\\ \frac{\lambda df}{f}=dx\\ \lambda\ln|f|=x+C. $$ Also, $$ f(0)=\int_{0}^{0}f(t)dt=0. $$

So, $\lambda\ln|0|= C$ which is impossible.

Answer 2

Because if $f$ is an eigenvector of that operator with an eigenvalue $\lambda\neq0$, then$$\int_0^xf(t)\,\mathrm dt=\lambda f(x)$$and therefore$$0=\int_0^0f(t)\,\mathrm dt=\lambda f(0),$$which implies that $f(0)=0$.