Using integration by parts to show that if a function integrates to zero against every function with zero mean, then it is constant We are given  
$$ \int_{0}^{1}f(x)v(x)dx = 0 $$ whenever $$ \int_{0}^{1}v(x)dx = 0 $$ with $f$ and $v$ smooth.
We are asked to use integration by parts to show $f$ is constant if we have the above condition.
We arrive at $$ \int_{0}^{1}f(x)v(x)dx = [f(x)\int v(x)dx]\bigg|_0^1 - \int_{0}^{1}((\int v(x)dx)f'(x))dx = 0 $$ ie. $$ \int_{0}^{1} fv = [f\int v]\bigg|_0^1 - \int_{0}^{1}((\int v)f') = 0 $$ 
How can we proceed from here to show $f$ is constant?
I have been told several things, such as that $[f\int v]\bigg|_0^1 = 0$, but I just get $f(1)V(1) - f(0)V(0) = (f(1)-f(0))c $ for some $c$ since $V(0)=V(1)$ by definition.
I've also been told that this integration by parts shows that $\int fv = -\int v \int f'$ but I don't see how that conclusion was arrived at.
 A: 
I have been told that $[f\int v]\big|_0^1=0$, but i just get that $f(1)V(1)-f(0)V(0)=(f(1)-f(0))c$ since $V(0)=V(1)$ be definition.

You are right, it will not always work out that $[f\int v]\big|_0^1=0$. However, when integrating by parts, you can choose any antiderivative $V$ of $v$. In particular, you can let $V(x)=\int_0^x v(t)\,dt$, in which case you do have $V(0)=V(1)=0$, so that $[fV]\big|_0^1$ vanishes. This leaves you with
$$
\int_0^1 Vf'\,dx = 0
$$
To complete the proof, for all $s\in (0,1)$, assume by way of contradiction that $f(s)\neq0$, and without loss of generality, $f'(s)>0$. Choose an interval $(s-\delta,s+\delta)$ where $f(x)>\epsilon$, for some $\epsilon>0$. Then, choose a function $v$ with zero mean so that $V=\int_0^x v(t)\,dt$ is a bump function on the interval $(s-\delta,s+\delta)$. This allows you to to show that $\int_0^1 Vf'dx>0$, contradicting the displayed equation, so we must have $f'$ is always zero.
A: You can prove this by contradiction. Suppose $f(x)$ is not constant on the entire interval $(0,1)$, i.e. there is an interval (a,b) contained in $(0,1)$ on which $f'(x) > 0$ (you may also use $< 0 $). Instead of defining a $v(x)$ let us work with its primitive function $V(x)$ that is continuously differentiable (i.e. the corresponding $v(x)$ is smooth as required). Take the function
\begin{align*}
V(x) = - \frac{1}{2} \cos \left(\frac{2 \pi}{a- b} (x-a) \right) + \frac{1}{2}, \quad x \in (a,b)
\end{align*} and $0$ outside.
Note that the function is smooth as the derivatives on $x = a$ and $x = b$ match. Obviously its derivative $v(x)$ satisfies the condition $\int_0^1 v(x) \mathrm{d}x = 0$.
We have $V(0) = V(1) = 0$ so you may forget about the first term in your expression.
As both $V(x)$ and $f'(x)$ are $> 0 $ on $(a,b)$ we have $\int_0^1 V(x) f'(x) \mathrm{d} x \neq 0$ leading to a contradiction.
$\implies f'(x) = 0$, Q.E.D.
