Finding fix point of an integral operator T I'm trying to find the fix point of the operator $T(f) = \int_0^x f(y) (1/2 + y/2)/ dy$. 
My approach is to put $Tf = f(x)$ and taking the derivative $d/dx$ of each side, which give me 
$$ f(x)(1/2 + x/2) = df/dx$$
which is a separable ODE, with the solution $ f(x) = C \exp(x^2/4+x/2)$, but when I'm checking $Tf = f$ (plugging in my f(x) in the integral) I instead get $$ Tf = f-1$$ 
due to the lower limit $x=0$ 
 A: Since $Tf = f$ as functions, in particular, $Tf(0) = f(0)$.  On one hand, $Tf(0) = \int_0^0 \cdots \,\mathrm{d}y = 0$.  On the other hand, $f(0) = C \exp(0) = C$, so $C = 0$, thus $f \equiv 0$.  This should gives the correct fixed point.
A: $f \equiv 0$ is indeed the unique fixed point.
Note that $T$ can be understood as a linear map on $C[0,1]$ equipped with the supremum norm.
We have:
$$\|Tf\|_\infty = \sup_{x \in [0,1]} \left|\int_0^xf(y) \left(\frac12 + \frac{y}2\right)\,dy\right| \le \|f\|_\infty \int_0^1 \left(\frac12 + \frac{y}2\right)\,dy = \frac34 \|f\|_\infty$$
so $\|T\| \le \frac34 < 1$ so $T$ is a contraction mapping. Since $C[0,1]$ is a Banach space with respect to the supremum norm, the Banach fixed point theorem states that there exists a unique fixed point for $T$. Linearity gives $T0 = 0$ so $0$ is the desired fixed point.
It is important to note that your approach assumes that the  fixed point $f$ is differentiable. This assumption is not needed, as the above argument requires only continuity of $f$.
