Injectivity of the operator $(Ax)(t)=\int_0 ^1 k(s,t) x(s)ds$

Solution 1:

An easy one is $k(x,t)=k(x-t)$, and the (1 periodic) fourier series of $K$ has only non zero coefficients.

Indeed, by linearity it is enough to show that $Ax=0$ implies $x=0$. Now, $x$ let us compute the fourier coefficients of $Ax$. We have

$$ \hat{Ax}(n)=\int_0^1 Ax e^{-i2\pi n x}{\rm d}x=\int_0^1\int_0^1 k(x-t)x(t)e^{-i2\pi n t}e^{-i2\pi n (x-t)} {\rm d}x {\rm d}t $$ (as $\exp(-i2\pi n t)\exp({-i2\pi n (x-t)})=\exp(-2\pi nx)$). Changing variables to $t=t$, $u=x-t$, we obtain $$ \hat{Ax}(n)=\int_0^1 \int_0^1k(u)\exp({-i2\pi n u})x(t)\exp(-i2\pi n t){\rm d}u{\rm d}t $$ and this last expression is just the product of two integrals, namely $$ \hat{Ax}(n)=\hat{k}(n)\hat{x}(n). $$ Therefore, if for all $n\in \mathbb{Z}$, $\hat{k}(n)\neq0$, then $Ax=0$ implies $\hat{x}(n)=0$ for all $n$. Then Parseval's Theorem shows that $\int_0^1 x^2(t)dt=0$, and therefore, as $x$ is continuous, $x\equiv0$.