Using second derivative to find a bound for the first derivative 
Let $f$ be a twice differentiable function on $\left[0,1\right]$ satisfying $f\left(0\right)=f\left(1\right)=0$. Additionally $\left|f''\left(x\right)\right|\leq1$ in $\left(0,1\right)$. Prove that $$\left|f'\left(x\right)\right|\le\frac{1}{2},\quad\forall x\in\left[0,1\right]$$

The hint we were given is to expand into a first order Taylor polynomial at the minimum of $f'$. So I tried doing that:
As $f'$ is differentiable, it is continuous, and attains a minimum at $\left[0,1\right]$. Thus we can denote $x_{0}$ as the minimum, and expending into a Taylor polynomial of the first order around it gives us, for some $c$ between $x$ and $x_0$. 
$$T_{x_{0}}\left(x\right)=f\left(x_{0}\right)+f'\left(x_{0}\right)\left(x-x_{0}\right)+\frac{f''\left(c\right)}{2}\left(x-x_{0}\right)^{2}$$
Now at $x=0$ we have
$$T_{x_{0}}\left(0\right)=f\left(x_{0}\right)-x_{0}f'\left(x_{0}\right)+x_{0}^{2}\frac{f''\left(c\right)}{2}=0$$
And at $x=1$ we have $$T_{x_{0}}\left(0\right)=f\left(x_{0}\right)+\left(1-x_{0}\right)f'\left(x_{0}\right)+\left(1-x_{0}\right)^{2}\frac{f''\left(c\right)}{2}=0$$
and I'm pretty much stuck here..
So I tried a different approach using Mean Value Theorem directly, by Rolle's I know the derivative is $0$ somewhere (as $f(0)=f(1)$), suppose at $x_0$, so by mean value theorem $\frac{f'(x)}{x-x_0} =f''(c)\leq1$ for some $c$ and $f'(x)<x-x_0$.
But using this approach as well I'm not sure how to proceed, as this gives me the desired propety only in $1/2$ environment around $x_0$...
Any help?
 A: You were on the right track, but, as I suggested, the hint should have been to expand about the point $x_0$ where $|f'(x_0)|$ is a maximum.
Fix any $x_0\in [0,1]$ and, using Taylor's Theorem, write 
$$f(x)=f(x_0) + f'(x_0)(x-x_0) + \frac12 f''(c)(x-x_0)^2\quad\text{for some $c$ between $x_0$ and $x$.}$$
Plugging in $x=0$ and $x=1$ respectively, we arrive at
$$f'(x_0) = \frac12\big(f''(c_1)x_0^2 - f''(c_2)(1-x_0)^2\big) \quad\text{for some $c_1$ and $c_2$}.$$
Therefore $|f'(x_0)|\le \dfrac12\big(x_0^2 + (1-x_0)^2\big) \le \dfrac12$, since $x_0\in [0,1]$ is arbitrary.
A: This is an equivalent statement:
Let $g:[0,1] \to \mathbb R$ be a continuous function on $[0,1]$. Assume that $g$ is differentiable on $(0,1)$,
$$ 
\int_0^1 g(x) dx = 0 \qquad \text{and} \qquad  \forall x \in (0,1) \quad \bigl| g'(x) \bigr| \leq 1. \tag{1}\label{1} 
$$
Then $|g(x)| \leq 1/2$ for all $x \in [0,1]$.
Proof. By the Mean value theorem, for all $x,y \in [0,1]$ such that $x \neq y$ there exists $c \in (0,1)$ such that 
$$
\frac{g(y) - g(x)}{y-x} = g'(c).
$$
By the second assumption in $\eqref{1}$, for all $x,y \in [0,1]$ such that $x \neq y\ $ we have 
$$
\left| \frac{g(y) - g(x)}{y-x} \right| \leq 1.
$$
Consequently, for all $x,y \in [0,1]$ we have 
$$
-|y-x| \leq g(y) - g(x) \leq |y-x|.
$$
Integrating the last inequality with respect to $x$ over $[0,1]$ and using the first assumption in $\eqref{1}$, we get
$$
-\int_0^1 |y-x| dx \leq g(y) \leq \int_0^1 |y-x| dx.
$$
Thus, for all $y \in [0,1]$ we have
$$
\bigl| g(y) \bigr| \leq \int_0^1\!\! |y-x| dx =\frac{1}{2}\bigl( y^2 + (1-y)^2 \bigr) = \Bigl(y-\frac{1}{2}\Bigr)^2 + \frac{1}{4} \leq \frac{1}{2}.
$$
