I have the following exercise:
"Find the values $\lambda\in\mathbb{C}\setminus\{0\}$ s.t. the integral equation $$ \lambda\phi(x)-\int_0^1 e^{x+t}\phi(t)dt=f(x) $$ admits a unique solution $\phi_f$ for every given $f\in L^2(0,1)$. Discuss the regularity of $\phi_f$ when $f\in C^0([0,1])$."
I know that I have to find the eigenvalues of T, where $T:L^2(0,1)\rightarrow L^2(0,1)$ defined as $$ T(u)(x)=\int_0^1 e^{x+t}u(t)dt. $$ Once I have found the eigenvalues $\bar{\lambda}_i$ of T I have that the equation admits a unique solution for every f whenever $\lambda\ne\bar{\lambda}_i$. I have also that T is compact and self-adjoint since it is an Hilbert-Schmidt operator with kernel which is symmetric and in $L^2((0,1)\times(0,1))$.
But I can't find the eigenvalues of T. I have tried to find for which $\lambda$, $T(u)(x)=\lambda u(x)$ admits a solution $u(x)\ne 0$ by observing that $$ T(u)(x)=e^x\int_0^1 e^tu(t)dt=Ce^x $$ but then I get that every $\lambda\ne 0$ is an eigenvalue, which is clearly wrong.
The only possible eigen fucntions are multiples of $e^{x}$. Now $T(e^{x})=\int_0^{1}e^{t+x} e^{t}dt$ so if you equate this with $\lambda e^{x}$ you get $\int_0^{1}e^{t+x} e^{t}dt=\lambda e^{x}$. This says $\int_0^{1}e^{t} e^{t}dt=\lambda$ so $\lambda =\frac 1 2 ({e^{2}-1})$. This is the only eigen value.