I am a bit confused with the existence/non existence of a solution to the following equation: $$\int_{-T}^{T} R(t,\tau) x(\tau) d\tau = y(t), \forall t\in [-T,T]$$ where $y(t)=C$ (a constant function), $x(\tau)$ is the unknown function, and $$R(t,\tau)=1-\frac{\left|{t-\tau}\right|}{a}$$ What bothers me is that I've proven that this equation has a solution, and on the contrary I've proven (by another way) that it doesn't! And I don't find the mistake. If you could help me find the mistake, I would appreciate it.
Proof that this equation does not have a solution (if $C\neq 0$):
By contradiction, suppose there is a solution to $\int_{-T}^{T} \left(1-\frac{\left|{t-\tau}\right|}{a}\right) x(\tau) d\tau = C, \forall t\in [-T,T]$. Take a derivative (of both sides) with respect to $t$ and get: $$\int_{-T}^{T} \frac{(-1)}{a}sgn(t-\tau)x(\tau)d\tau=0, \forall t\in [-T,T]. $$ Which can be rewritten as: $$\frac{(-1)}{a}\int_{-T}^t x(\tau)d\tau + \frac{1}{a}\int_{t}^T x(\tau)d\tau=0, \forall t\in [-T,T]$$ Take another derivative with respect to $t$ to get: $\frac{(-2)}{a}x(t)=0$, which means that $x$ is the zero function. But that solution does not satisfy the equation. $$$$
Proof that this equation has a solution:
The operator $x\longmapsto \int_{-T}^{T} R(t,\tau) x(\tau) d\tau$ is a compact $\left[ \text{Fredholm's theory}\right]$ and self-adjoint (since $R(t,\tau)=R(\tau,t)$), and therefore there is an orthonormal basis $\left\{\psi_k(t)\right\}$ of eigenvectors of this operator. Denote the eigenvalues by $\left\{\lambda_k\right\}$. There is no eigenvalue $\lambda=0$ because the only solution to the equation $\int_{-T}^{T} R(t,\tau) x(\tau) d\tau=0$ is the zero function $x(t)=0$ (see the proof above for the non-existance of solution for $y(t)=C\neq 0$).
As $\left\{\psi_k(t)\right\}$ is an orthonormal basis, we can expand every function $y(t)$ (even the constant function $y(t)=C$) to $y(t)=\sum_{k=1}^{\infty} \beta_k \psi_k(t)$ where $\beta_k=\int_{-T}^{T} y(t) \psi_k^*(t) dt$. The following function is then the solution of the equation:
$$x(t)=\sum_{k=1}^{\infty} \frac{\beta_k}{\lambda_k} \psi_k(t)$$
To see that we substitute it: $$\int_{-T}^{T} R(t,\tau)x(\tau)d\tau = \int_{-T}^{T} R(t,\tau)\sum_{k=1}^{\infty} \frac{\beta_k}{\lambda_k} \psi_k(\tau)d\tau = \sum_{k=1}^{\infty} \frac{\beta_k}{\lambda_k} \int_{-T}^{T} R(t,\tau)\psi_k(\tau)d\tau = \sum_{k=1}^{\infty} \frac{\beta_k}{\lambda_k} \lambda_k \psi_k(t) = y(t)$$
Now, there are many places where my proof is not precise, for example, I am not sure that the series I gave for $x(t)$ converge, but what is the $real$ reason for one of these proofs to fail? And which one is wrong?