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.
 A: Assume that you can prove the statement for $0\leq\alpha\leq 1/2$ for all functions, and let us conclude that it follows for the rest of $\alpha$.
Let $\phi$ be a given function with continuous derivative on $[0,1]$ and $\int_0^1\phi(x)\,dx=0$, and let $\alpha>1/2$. We need to show that
$$\left|\int_0^{\alpha}\phi(x)dx\right| \leq \frac{1}{8}\max_{0\leq x\leq 1}|\phi'(x)|.$$
Let $\psi(x)=-\phi(1-x)$ (thus $\psi$ also has a continuous derivative on $[0,1]$ and $\int_0^1\psi(x)\,dx=0$). 
Since $1-\alpha\leq1/2$, we know that
$$\left|\int_0^{1-\alpha}\psi(x)dx\right| \leq \frac{1}{8}\max_{0\leq x\leq 1}|\psi'(x)|.$$
But since $\psi'(x)=\phi'(1-x)$,
$$
\frac{1}{8}\max_{0\leq x\leq 1}|\phi'(x)|
=\frac{1}{8}\max_{0\leq x\leq 1}|\psi'(x)|.$$
Also, the change of variables $x\mapsto 1-x$, and the property $\int_0^1\phi(x)\,dx=0$ gives
$$
\int_0^{1-\alpha}\psi(x)\,dx=-\int_\alpha^{1}\phi(x)\,dx=\int_0^\alpha\phi(x)\,dx.
$$
Thus,
$$\left|\int_0^{\alpha}\phi(x)dx\right| =\left|\int_0^{1-\alpha}\psi(x)dx\right| \leq \frac{1}{8}\max_{0\leq x\leq 1}|\psi'(x)|= \frac{1}{8}\max_{0\leq x\leq 1}|\phi'(x)|.$$
A: 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 $\frac{1}{2}$).
