Symmetry argument and WLOG Suppose that $f:[0,1]\rightarrow \mathbb{R}$ has a continuous derivative and that $\int_0^1 f(x)dx=0$. Prove that for every $\alpha \in (0,1)$, 
$$\left|\int_0^{\alpha}f(x)dx\right| \leq \frac{1}{8}\max_{0\leq x\leq 1}|f'(x)|.$$
The solution goes like this: Let $g(x)=\int_0^x f(y)dy$, then $g(0)=g(1)=0$, so the maximum value of $|g(x)|$ must occur at a critical point $y\in (0,1)$ satisfying $g'(y)=f(y)=0$. We may take $\alpha = y$ hereafter.
Since $\int_0^{\alpha}f(x)dx= - \int_0^{1-\alpha}f(1-x)dx$, we may without loss of generality assume that $\alpha \leq 1/2$.
I don't understnd why the WLOG argument works in this case. Could someone please explains? Thanks in advance.

Consider an $\alpha$ that is greater than $\frac{1}{2}$. Then you know that $1-\alpha \leq \frac{1}{2}$ and then we can make the substitution they give you to "pretend" we have the kind of $\alpha$ that we want (Less than or equal to $\frac{1}{2}$).
