Is there a solution to this integral equation? The problem is related to this question: How to find eigenfunctions
of a linear operator (follow-up
question) I posted earlier.
Suppose I want to solve the following integral equation:
$$\int_0^1 K(x,t)y(t)dt=\sqrt{2x^2-2x+1}$$ where
$$K(x,t)=\max((1-x)t,(1-t)x),0<x<1,0<t<1.$$ 
Eigenfunctions of $ K(x,t)$ was found by @oenamen in the answer to the
above-mentioned question.  I thought one should be able to use the
eigenfunctions to find a solution to the above equation but I fail to
see if this is the case. I would appreciate any suggestions or
comments on this.
 A: Explicit solution
Take the second derivative of each side of the integral equation,
$$y = f''.$$
Plugging this back into the integral equation we find that $f$ must satisfy the Robin boundary conditions obeyed by the eigenfunctions,
$$\begin{eqnarray*}
f'(0) + f(0) &=& 0 \\
f'(1) - f(1) &=& 0.
\end{eqnarray*}$$
(We should expect to be able to expand a function obeying these boundary conditions in terms of the eigenfunctions.)
Since $f(x) = \sqrt{2x^2-2x+1}$ satisfies these boundary conditions the solution exists.
Thus,
$$y = \frac{1}{(2x^2-2x+1)^{3/2}}.$$
Eigenfunction expansion 
We write the integral equation schematically as
$$f = K y.$$
Let $y_n$ denote the $n$th eigenfunction of $K$,
$$y_n = \lambda_n K y_n.$$
These have been found for the operator pertinent to this question here.
The eigenfunctions are orthogonal and we assume they have been normalized
$$y_m \cdot y_n = \delta_{mn}.$$
(The inner product is $f\cdot g = \int_0^1 d t\, f(t)g(t)$.)
Picard's theorem mentioned in the comments states that $f$ can be expanded in terms of the eigenfunctions,
$$f = \sum f_n y_n$$
where $f_n = y_n \cdot f$.
Then
$$\begin{eqnarray*}
y &=& K^{-1} \sum f_n y_n \\
&=& \sum \lambda_n f_n y_n \\
&=& \sum c_n y_n
\end{eqnarray*}$$
Thus, the coefficients of the expansion for $y$ are $c_n = \lambda_n f_n$.
Some details 
Define
$$y_0 = A_0(\sqrt{\lambda_0} \cosh\sqrt{\lambda_0} x - \sinh\sqrt{\lambda_0} x)$$
and
$$y_n = A_n(\sqrt{\mu_n} \cos\sqrt{\mu_n} x - \sin\sqrt{\mu_n} x)$$
for $n\ge 1$.
Note that $\lambda_n = -\mu_n$ for $n\ge 1$.
Normalizing we find
$$A_0 \approx 0.769,\hspace{3ex}
A_1 \approx 0.672, \hspace{3ex}
A_2 \approx 0.241, \hspace{3ex}
A_3 \approx 0.154, \hspace{3ex} \ldots$$
The coefficients $c_n = \lambda_n \int_0^1 d t\, y_n(t) f(t)$ are
$$c_0 \approx 1.94, \hspace{3ex}
c_1 \approx 0, \hspace{3ex}
c_2 \approx -0.757, \hspace{3ex}
c_3 \approx 0, \hspace{3ex}\ldots $$
The function $y = \sum_{n=0}^3 c_n y_n$ already provides a very good approximate solution to the integral equation. 
