Show that $\ker(C)=\{0\}$ - what is wrong with this arguement?

Question

Consider the operator $C:X\to X$ where $X=C([0,1])$ and $y\in X$ where $Cy(t)=\int_0^ty(\tau)d\tau+y(t)$. I want to show that $\ker(C)=\{0\}$. Here is my arguement, but I am not convinced that it is a very good one - what is wrong with it?

Suppose that, $$Cy(t)=0$$

$$\implies\int_0^ty(\tau)d\tau+y(t)=0$$ $$\implies\int_0^ty(\tau)d\tau=-y(t)$$ $$\implies\int_0^1\int_0^ty(\tau)d\tau dt=-\int_0^1y(t)dt $$ $$\implies\int_0^ty(\tau)d\tau=-\int_0^1y(t)dt$$

Now, the upper limit $t$ in the left integral is such that it must be from the interval $(0,1]$. Without loss of generality, suppose that this upper limit is $t=1$. Then,

$$\implies\int_0^1y(\tau)d\tau=-\int_0^1y(t)dt$$

$$\implies y(t)\equiv0,\,\forall \, t\in(0,1]$$

As $y(t)\equiv0$ is the only function that could satisfy this equality. What is wrong with this arguement? Is it valid? I have doubts that it is...

Answer 1

There are precisely two errors in your reasoning (each of which was separately observed in the comments on your question):

The first error is that the passage from $$\int_0^1\int_0^ty(\tau)d\tau dt=-\int_0^1y(t)dt $$ to $$\int_0^ty(\tau)d\tau=-\int_0^1y(t)dt$$ is completely unjustified. (Why did you drop the $\int_0^1 ... dt$ on the left? Are you assuming the integrand ($\int_0^ty(\tau)d\tau$) is constant with respect to the variable $t$? It is not.)

The second error is that $$\int_0^1y(\tau)d\tau=-\int_0^1y(t)dt$$ only implies that $\int_0^1y(t)dt = 0$; it does not follow that $y$ is itself constantly $0$.

All the rest of your reasoning would have been sound, but each of these steps is unsupported and in fact fallacious.

Answer 2

Arthur pointed out the main flaw in your argument. A possible argument goes like this. (thanks to Sridhar Ramesh for pointing out a mistake in my original argument)

The start of your argument is fine: take a function $y$ in Ker$(C)$, then $$ \int_0^t y(\tau)d\tau = - y(t) . $$ The left hand side of this equation is differentiable in $]0, 1[$, so the right hand side must also be differentiable in $]0, 1[$. Taking the derivative with respect to $t$ on both sides yields $$ y(t) = -y'(t) $$ for all $t \in ]0, 1[$. Recall from an elementary course in differential equations that the solutions to this equation are exactly of the form $$ y(t) = Ke^{-t} \quad K \in \mathbb{R}. $$ So the only possible elements of Ker$(C)$ must be of the form $Ke^{-t}$. On the other hand, if we plug in such a function into $C$ we get $$ Cy(t) = \int_0^t K e^{-\tau}d\tau + Ke^{-t} = K(1 - e^{-t}) + Ke^{-t} = K $$ which is $0$ if and only if $K = 0$. Hence, Ker$(C)$ = $\lbrace 0 \rbrace$.