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Let $X=C([0,1],\mathbb{R})$ (equipped with the supremum norm). Let $A$ be the operator defined for each $x\in X$ by $$(Ax)(t)=\int_0 ^1 k(s,t) x(s)ds,$$ where $k:[0,1]\times [0,1]\to \mathbb{R} $ is continuous.

Can we find a sufficient condition on $k$ to make $A$ injective ?

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    $\begingroup$ A sufficient condition is that the family of functions $\{s\mapsto k(s,t): t\in [0,1]\}$ has dense linear span in $L^2[0,1]$. I imagine this condition is not easy to check, though. $\endgroup$ – user147263 May 25 '14 at 4:46
  • $\begingroup$ Not an answer but maybe useful: A sufficient condition surely is that $k$ be a Green's function (typically denoted by $G$). (Related to the discussion we had in the other question of yours). I think that Green's functions can be characterized as the functions that decompose as follows: $$ G(x, y) = \sum_{k=1}^{\infty} \mu_k \Phi_j (x) \Phi_j(y),$$ where $\mu_j$ is real and tends to $0$ and $\Phi_j$ is a complete orthonormal system consisting of smooth functions. I am not entirely sure, though. Moreover, this is probably not a real answer $\endgroup$ – Giuseppe Negro May 26 '14 at 15:29
  • $\begingroup$ Fourier Kernel should be another example. $$k(s,t) = \exp(j2\pi st)$$ $\endgroup$ – Yan Zhu Jun 16 '14 at 8:52
  • $\begingroup$ for a related MO post see mathoverflow.net/questions/243425/… $\endgroup$ – Ali Taghavi Feb 26 '17 at 16:09
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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, $x\equiv0$.

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  • $\begingroup$ Can you add some details please ? $\endgroup$ – user146010 Jun 20 '14 at 9:14
  • $\begingroup$ @Edwin I edited the question to include the details $\endgroup$ – username Jun 20 '14 at 11:59

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