If $f$ is continuous and $f(x)=\int_{0}^{x} f(t)dt$, prove $f=0$ without FTC. Small issue with solution.

Question

If $f$ is continuous and $f(x)=\int_{0}^{x} f(t)dt$, prove $f=0$ without FTC.

Question is from Courant Intro to Calculus any Analysis I, s2.3-3. It comes before fundamental theorem of calculus (FTC). This is solution:

How do we know that $\lim\limits_{n\to\infty} x^n$?

Answer 1

We do indeed have $\lim_{n \to \infty} x^n = 0$ only for $x \in (-1,1)$. This implies that $f(x) = 0 \forall x \in (-1,1)$. But since $f$ is continuous this implies $f(x) = 0$ also for $x=-1$ and $x=1$.

Answer 2

Also you can have this alternative solution without applying FTC. Consider $\psi(x)=\int_{0} ^{x} f(t)dt$, then $\psi$ is differentiable at each $x\in \mathbb{R}.$ Therefore $f$ is also differentiable and $f'(x)=f(x)$ also $f(0)=0$. Now consider $\phi(x)=f(x) \exp(-x)$, then $\phi ' =0 ,$ therefore $\phi(x) = c,$ a constant, but $\phi(0)=0$. This implies $f(x)=0. \psi $ is differentiable at $x_0$ because $|\frac{\psi(x)-\psi(y)}{x-y}-f(x_0)|=|\frac{1}{x-y}\int_x ^y[f(u)-f(x_0)]du |< \epsilon$ for any $x_0-\delta<x \leq x_0 \leq y < x_0+\delta .$