$g(x)$ is defined as the following : $$g(x)= \int_0^1 \! K(x,y) f(y)\ \mathrm{d}x $$
where $K(x,y)$ is continuous in $ 0 \leq x \leq 1 $ , $ 0 \leq y \leq 1 $, and $f(x)$ is continuous in $ 0 \leq x \leq 1 $. When $g(x)$ satisfies the following :
$$ \frac{d^3g}{dx^3} = f \space \space \space \space \space \space \space \space \space \space \space \space \space \space (0 \leq x \leq 1) $$ $$ g(0)=g(1)=\frac{dg(0)}{dx}= 0 $$
How do I show that $K(x,y)$ does not depend on $f(x)$? Moreover, how do I show $\frac{\partial}{\partial x}K(x,y)$ is continuous and that $\frac{\partial^3}{\partial x^3}K(x,y) = 0$ ?
Two thirds of the way:
Forget about $K$ for the moment. Assume we are given a continuous $f:\ [0,1]\to{\mathbb R}$, and we want to construct a $g$ fulfilling the given conditions.
From $g'''=f$ we conclude that $$g''(x)=\int_0^x f(t)\>dt +c\ ,$$ where $c$ is as yet undetermined. This leads to $$g'(x)=g'(0)+\int_0^x g''(t)\>dt=\int_0^x\left(\int_0^t f(\tau)\>d\tau + c\right)\>dt\ .$$ Partial integration gives $$g'(x)=\int_0^x (x-t)f(t)\ dt+c x\ .$$ This in turn implies $$g(x)=g(0)+\int_0^x g'(t)\>dt=c{x^2\over2}+\int_0^x\left(\int_0^t (t-\tau)f(\tau)\>d\tau\right)\>dt\ .$$ Now do another partial integration, fix $c$ such that $g(1)=0$, and you can read off the requested $K$.