Solving Fredholm integral equation in Lp

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!

• in fact the last "another idea" is correct, so you already answered! – Pietro Majer Apr 23 at 10:08
• @PietroMajer , thank you so much for your valuable comment! :) – jonathan wolf Apr 24 at 17:06
• 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. – Pietro Majer Apr 25 at 9:20