Why $f(x)=e^x$ is a fixed point of $T$?

Question

Let $(C[0, 1], d_∞)$ be the metric space of continuous functions on $[0, 1]$ , where the distance function is defined by $$d_∞(f,g)= \sup_{x\in [0,1]}|f(x)-g(x)|$$

Consider the function $T:(C[0, 1], d_∞)\rightarrow (C[0, 1], d_∞)$ defined by

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

Why do we have $f(x)=e^x$ as a fixed point of $T$?

I am so confused because $T(e^x)=e^x-1$. But, $e^x \neq e^x-1$.

Answer 1

It is not. The only fixed point is $0$. $Tf=f$ implies $f'(x)=f(x)$ so $f(x)=ce^{x}$ for some constant $c$. But $Tf(0)=0$ so we must have $c=0$.