Kernel of $f\mapsto f + \int f$

Question

I have the following linear opearator

$$O: C[0,1] \to C[0,1], \ \ \ Of(x) = f(x) + \int_{0}^{x} f(s)\, ds,\ x \in [0,1].$$

I want to prove that that $\ker(O)=\{0\}$. I tried to match the function to zero.

The solution of the differential equation may be $$y(x)=Ce^{-t},\:\; C \in \mathbb{R}.$$

Anyone can help here or give a hint?

Answer 1

Let $f \in C[0,1]$ with $Of = 0$. Then \begin{equation} f(x) + \int_0^x f(s) \text{ }ds = 0 \Leftrightarrow f(x) = - \int_0^x f(s) \text{ }ds \end{equation} for all $x \in [0,1]$. If you don't see it by now, you may differentiate and obtain $f' = -f$ leading to the unqiue solution $f(x) = f(0)e^{-x}, x \in [0,1].$ Since $0 = (Of)(0) = f(0)$, we may conclude $f(x) = 0$ for all $x \in [0,1]$.

Answer 2

No need for any fancy differential equation stuff.

Suppose $Of=0$, then $f(x) = - \int_0^x f(t) dt$ and so $|f(x)| \le x \|f\|$. Using the equality again we get $|f(x)| \le \int_0^x t \|f\| dt = {x^2 \over 2} \|f\|$ which gives $\|f\| \le {1 \over 2} \|f\|$ and hence $f=0$.