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Let $k:[0,1]^2\rightarrow (0,+\infty)$ be a continuous function and let $f,g:[0,1]\rightarrow (0,+\infty)$ be measurable functions. We assume that $$\forall x\in [0,1],\quad f(x)=\int_0^1 k(x,y) g(y) dy,\hskip15pt g(x)=\int_0^1 k(x,y) f(y) dy. $$ We want to prove that $f=g$.

Note that defining the operator $K$ by $(Ku)(x)=\int_0^1 k(x,y) u(y) dy$, this amounts to proving that the assumption $f=K^2f$ implies $f=Kf$ for a positive $f$. In other words, if $(Id-K^2)f=0$, it should imply $(Id-K)f=0$ whenever $f$ is positive. If the operator $Id+K$ were one-to-one, it would imply the sought result, but I do not see why this invertibility property should hold.

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  • $\begingroup$ Something wrong with my answer? $\endgroup$
    – username
    Apr 19 at 19:49
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This is related to Leray's Eigenvalue Problem . Under your hypothesis, there is at least one positive eigenvalue, with a corresponding non negative eigenvector $u$ such that $$ Ku = \lambda u, $$ by Arzela-Ascoli's Theorem and Brouwer's Fixed point applied to $$ F(u)(x)= \frac{\int_0^1 k(x,y) u(y) dy}{\int_0^1\int_0^1 k(t,y) u(y) dy dt}. $$ Now, you have $K^2$ a positive, compact, linear operator acting on $L^{\ldots}(0,1)$ such that both $$ K^2(f)=f \mbox{ and } K^2(u) = \lambda^2 u, $$ that is, two stricly positive eigenvalues with positive eigenvectors. By the uniqueness property of such eigenpairs (Perron-Frobenius / Krein-Rutman Theorem) $\lambda=1$, and $f=u$, so indeed, $Kf=f$.

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