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?

score 6 · Answer 1 · answered Feb 08 '18 at 15:36

6

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 $f=0$.

answered Feb 08 '18 at 15:36

copper.hat

172,524

I think u meant $f(x) = - \int_0^x f(s) ds$ and $|f(x)| \le \int_0^x s |f| ds = {x^2 \over 2} |f|$ – Johnny_Vz Feb 11 '18 at 12:29
@Johnny_Vz: No, I meant to write $t$. In any event, the particular variable used is irrelevant other than not being $x$. – copper.hat Feb 11 '18 at 16:31

score 4 · Accepted Answer · answered Feb 08 '18 at 14:42

4

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 $x \in [0,1]$.

answered Feb 08 '18 at 14:42

Viertel

416
4
10

Actually, differentiation is not permitted here since you're working on $C[0,1]$ rather than $C^1[0,1]$, isn't it? – ComplexF Feb 08 '18 at 15:41
1

It is permitted since the right side of the equivalence tells you that f ist differentiable. – Viertel Feb 08 '18 at 16:20
in this case it is permitted – Johnny_Vz Feb 08 '18 at 16:30
Of course, you're completely right, sorry 'bout that... – ComplexF Feb 08 '18 at 16:47

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

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