Let $K:L^2[0,1]^{d_1}\to L^2[0,1]^{d_2}$ be integral operator $$(Kf)(y) = \int f(x)k(x,y)d x.$$ If $d_1>d_2$ is it possible for $K$ to be injective?, e.g. let's take $d_1=2,d_2=1$.

More generally, does the injectivity of $K$ impose any restrictions on $d_1,d_2$.

  • $\begingroup$ Is there any information on $k(x,y)$? For instance, in the trivial case when $k(x,y)$ does not depend on $y$ and vanishes for all $x \in (0,0.5)$, then $Kf(y) =0 \ \forall y$ for some non-zero function $f$, so the operator is not injective. $\endgroup$ – Yuxin Wang Nov 29 '18 at 1:43
  • $\begingroup$ Thank you. I don't have any conditions. I need an example of injective $K$ when $d_1>d_2$ (or at least in knowing whether this is possible). I'm not interested in counterexamples. $\endgroup$ – Lionville Nov 29 '18 at 1:45
  • $\begingroup$ Well, what are your thoughts on the question so far? $\endgroup$ – jgon Nov 29 '18 at 2:30

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