Proving that if $|f''(x)| \le A$ then $|f'(x)| \le A/2$

Suppose that $f(x)$ is differentiable on $[0,1]$ and $f(0) = f(1) = 0$. It is also known that $|f''(x)| \le A$ for every $x \in (0,1)$. Prove that $|f'(x)| \le A/2$ for every $x \in [0,1]$.

I'll explain what I did so far. First using Rolle's theorem, there is some point $c \in [0,1]$ so $f'(c) = 0$.

EDIT: My first preliminary solution was wrong so I tried something else. EDIT2: Another revision :\

I define a Taylor series of a second order around the point $1$: $$f(x) = f(1) + f'(1)(x-1) + \frac12 f''(d_1)(x-1)^2$$ $$f(0) = f(1) + f'(1)(-1) + \frac12 f''(d_1)(-1)^2$$ $$|f'(1)| = \frac12 |f''(d_1)| <= \frac12 A$$

Now I develop a Taylor series of a first order for $f'(x)$ around $1$: $$f'(x) = f'(1) + f''(d_2)(x-1)$$ $$|f'(x)| = |f'(1)| + x*|f''(d_2)|-|f''(d_2)| \leq \frac{A}{2} + A - A = \frac{A}{2}$$

It looks correct to me, what do you guys think?

Note: I cannot use integrals, because we have not covered them yet.

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$c$ does not need to be $1/2$. –  Andres Caicedo Jan 9 at 18:38
Yeah I thought about that, but I tried drawing a sample function graph a number of times, and it seemed like it had to be going through the middle if its uniformly continuous. Perhaps I misunderstood the concept of uniform continuity? Does it even play a role here? –  Xsy Jan 9 at 22:18
I've edited the question. My first solution was wrong so I tried something else, would love to know if its correct. –  Xsy Jan 12 at 0:52
I think it's alright. Don't forget to mention about $d$ that it is between $c$ and $x$. –  GinKin Jan 12 at 17:54
The new edit isn't right either. Since $x<1$, the right expression is $$|f'(x)|=|f'(1)|+|f''(d_2)|(1-x)\le\frac A2+A(1-x),$$ and this is not enough to conclude what you want. –  Andres Caicedo Jan 12 at 18:15

I've already give an answer without Integral to this question in another post. I'll duplicate the answer here. If it's against policy please tell me what I should do.

I manage an answer using both time a taylor expansion center at a point $x \in \left [0, 1 \right ]$. I get $$f(h) = f(x)+f'(x)(h-x) + \frac{1}{2} f''(\xi(h))(h-x)^2$$

now for h = 0 and h = 1, I get

$$0=f(x)-xf'(x)+\frac{1}{2}x^2f''(\xi(0))$$ and $$0=f(x)+f'(x)-xf'(x)+\frac{1}{2}(1-x)^2f''(\xi(1)).$$ Subtracting one to the other to get $f(x)$ out and a little manipulation yield: $$|f'(x)| = \frac{1}{2}|x^2f''(\xi(0))-(1-x)^2f''(\xi(1))|\leq\frac{1}{2}(|x^2f''(\xi(0))|+|(1-x)^2f''(\xi(1))|)\leq \frac{A}{2}(|x^2|+|1-x|^2)$$

since $x \in \left [0, 1 \right ], (|x^2|+|1-x|^2) \leq1$ and we get the result.

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If $f''$ exists on $[0,1]$ and $|f''(x)| \le A$, then $f'$ is Lipschitz continuous, which implies that $f'$ is absolutely continuous. So $f'(a)-f'(b)=\int_{a}^{b}f''(t)\,dt$ for any $0 \le a, b \le 1$. And, of course, the same is true of $f$ because $f$ is continuously differentiable.

Because $f(0)=0$, one has the following for $0 \le x \le 1$: $$f(x) = \int_{0}^{x}f'(t)\,dt = tf'(t)|_{0}^{x}-\int_{0}^{x}tf''(t)\,dt=xf'(x)-\int_{0}^{x}tf''(t)\,dt.$$ Similarly, because $f(1)=0$, one has the following for $0 \le x \le 1$: $$f(x)=\int_{1}^{x}f'(t)\,dt = (t-1)f'(t)|_{1}^{x}-\int_{1}^{x}(t-1)f''(t)\,dt=(x-1)f'(x)-\int_{1}^{x}(t-1)f''(t)\,dt.$$ Subtracting the second equation from the first gives $$0=f'(x)-\int_{0}^{x}tf''(t)\,dt+\int_{1}^{x}(t-1)f''(t)\,dt.$$ Therefore, $$|f'(x)| \le A\left(\int_{0}^{x}t\,dt + \int_{x}^{1}(1-t)\,dt\right)=\frac{A}{2}(x^{2}+(1-x)^{2}).$$ The maximum of the expression on the right occurs at $x=0$ and $x=1$, with a value of $A/2$. Therefore, $|f'(x)| \le A/2$.

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Thank's for the solution but we haven't covered integrals yet, so I can't really use/understand your solution at the moment. –  Xsy Jan 10 at 11:43