While doing my study on the boundary-crossing time of a stochastic process, I happened to deal with the following question which is somehow related to Fredholm theory.


**Question** : Suppose $K$ is continuous non-negative function on $[0,T]^2$ such that $K(t,s)= 0 $ if and only if $s \ge t$. Prove or disprove the following statement:


"If $f$ is integrable function on $[0,T]$, then $ \int\_{[0,T]} K(t,s)f(s)ds= 0 \hspace{3mm} \forall t \iff f= 0 $ a.s"


**Thought** : This statement is reasonable because for each $t$, $K(t, \cdot)$ can mimic the role of the indicator function $\mathbb{1}\_{[0,t]}$ . 


I'm essentially looking for an answer helping me proceed with that question, for example, an approach, or a condition to be added in order to make the question more agreeable,etc. Nevertheless, no matter how your help is, I always deeply appreciate it.


Thank you.



Injectivity of a Fredholm operator
Surprisingly (to me), the statement is false. My counterexample is a little messy, but the idea is fairly simple.


Take $T = 1$ and set $a\_n = \frac{1}{n}$ and $b\_n = 1 - \frac{1}{n}$ for $n \in \mathbb{N}$. Define $f$ by setting $f(t) = 1$ on the intervals $[a\_{2n+2}, a\_{2n+1})$ and $f(t) = -1$ on the intervals $[a\_{2n+1},a\_{2n})$. The point of using $\frac{1}{n}$ is that the lengths of adjacent intervals are approximately equal for large $n$ (i.e., their ratio goes to $1$). We seek a family of functions $k\_t(s)$ which is continuous in both $s$ and $t$, which satisfies $k\_t^{-1}(\{0\}) = [t, 1]$, and such that $\int k\_t(s)f(s) = 0$ for all $t$.


First we construct $k\_1$. We impose the condition that $k\_1(a\_{2n}) = b\_{2n}$ for all $n$. Then we define $k\_1$ on $[a\_{2n+2}, a\_{2n}]$ to be any continuous function which satisfies $\int\_{a\_{2n+2}}^{a\_{2n}} k\_1f = 0$ (i.e., $\int\_{a\_{2n+2}}^{a\_{2n+1}} k\_1 = \int\_{a\_{2n+1}}^{a\_{2n}} k\_1$) and satisfies $b\_{2n} \leq k\_1 \leq \alpha\_nb\_{2n}$ where $\alpha\_n \to 1$ as $n \to \infty$. (Possible because successive intervals have approximately the same length.) Then $k\_1$ is continuous, its integral against $f$ is zero, and $k\_1^{-1}(\{0\}) = \{1\}$.


Next, for each $n \geq 1$ let $k\_{a\_{2n}}(s) = k\_1(s) - b\_{2n}$ for $s \in [0,a\_{2n}]$ and let it be $0$ for $s > a\_{2n}$. These functions also have the desired properties.


To complete the construction I have to show that you can continuously interpolate $k\_{a\_{2n}}$ and $k\_{a\_{2n + 2}}$ while preserving the desired properties. You can do this by letting $k\_t(s) = k\_1(s) - (1-t)$ for $s \in [0,a\_{2n+4}]$ (this ensures that $\int\_0^{a\_{2n+4}} k\_t f = 0$) and dealing with the $s \in [a\_{2n+4},a\_{2n}]$ case separately. On the latter interval I don't have a simple formula, but I think it's clear that this can be done. If I had to write down an explicit formula I'd make each $k\_t$ piecewise linear so you're just varying a couple of values of $k\_t$ within $[a\_{2n+4},a\_{2n}]$ in a way that keeps the integral against $f$ zero and makes $k\_t$ to go zero at $s = t$.


Well, that's a sketch of a counterexample.


**tl;dr**: for a counterexample to work, $f$ must take both positive and negative values in any neighborhood of $0$. Once such an $f$ is chosen, find functions $k\_{1/n}$ which are continuous, nonzero on $[0,1/n)$ and zero on $[1/n,1]$ and whose integral against $f$ is zero. Finally, continuously interpolate between $k\_{1/n}$ and $k\_{1/(n+1)}$ to get functions $k\_t$ with the same properties. You have to be a little careful in the last step, but there's no serious obstruction.


