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I have a very simple integral equation $$ f(x) - \lambda \int_a^be^{x-y}f(y)dy=1 $$ which is Fredholm of the 2nd kind, with separable kernel. It is needed to find the values of $\lambda\in\mathbb R$, for which this equations has solutions in $L_p(a, b)$ with $1\leq p\leq \infty$.

My attempt at solution is as follows: due to the separability of the kernel, we can write $$ f(x) - \lambda e^x\underbrace{\int_a^be^{-y}f(y)dy}_{\equiv \alpha}=1 \quad \Rightarrow\quad f(x) = 1+\alpha e^x, $$ $$ 1+\alpha e^x - \lambda e^x\int_a^be^{-y}\left[1+\alpha e^y\right]dy=1, $$ $$ \alpha = \lambda \int_a^be^{-y}\left[1+\alpha e^y\right]dy = \lambda\left[(e^{-a}-e^{-b})+\alpha(b-a)\right], $$ from where it follows that $$ \alpha = \frac{\lambda (e^{-a}-e^{-b})}{ \lambda(a -b) +1}, \quad \lambda\neq\frac{1}{b-a} $$ hence, $f(x)$ has the following form: $$ f(x) = 1 + \frac{\lambda (e^{-a}-e^{-b})}{ \lambda(a -b) +1} \, e^{x} $$

And after this step I'm stuck. It is needed to find the values of $\lambda$ for which $f(x)\in L_p(a,b)$, with $1\leq p\leq \infty$. Since I'm not very good in functional analysis, the only thing that comes to my mind is to consider the definition of $L_p$-norm. But that doesn't seem to lead anywhere.

Another idea is that $f(x)$ as written above is $C[a, b]$, so that it is perhaps (?) $L_p$ over the compact $[a, b]$, for any $\lambda\neq\frac{1}{b-a}$. But I may be wrong here.

Thank you for any hint!

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    $\begingroup$ in fact the last "another idea" is correct, so you already answered! $\endgroup$ – Pietro Majer Apr 23 at 10:08
  • $\begingroup$ @PietroMajer , thank you so much for your valuable comment! :) $\endgroup$ – jonathan wolf Apr 24 at 17:06
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    $\begingroup$ To put it in a slightly more general frame: your operator is a rank-one perturbation of the identity (hence Fredholm-$0$): $f\mapsto f-\lambda \langle\phi,f\rangle u$. Here with $u(x):=e^x$ and $\langle\phi,f\rangle:=\int_a^b e^{-y}f(y)dy$; the same conclusions hold in general. $\endgroup$ – Pietro Majer Apr 25 at 9:20

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