Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can't seem to get this subject very well.

Let $f(x)$ be twice differentiable on $[0,1]$, and that there is a constant $A$ so that $|f''(x)|\le A$. Show that if $f(0)=f(1)=0$, then $|f'(x)|\le {A\over2}$ for all $x\in[0,1]$.

Thanks in advance for any help. Would prefer hints please for my learning.


share|cite|improve this question
Do you see intuitively why this should be true? That'll be the first step to finding a proof. – user108903 Jan 22 '13 at 19:15
up vote 1 down vote accepted

for $x \in (0,1]$ : use taylors thrm:

$$f(0) = f(x-x ) =f(x) -xf^\prime(x) + \frac{x^2}{2}f^{\prime\prime}(x - h_1x).....(1)$$

put $x=1$ in $(1) \implies |f^\prime (1)| \leq \frac{A}{2}$

for $x \in[0,1)$ use taylors thrm.

$$f(1) =f(x+(1-x)) = f(x) + (1-x)f^\prime(x) +\frac{(1-x)^2}{2}f^{\prime\prime}(x +h_2(1-x)) .....(2)$$

put $x=0$ in $(2)\implies |f^\prime (0)| \leq \frac{A}{2}$

now $$(2) - (1) \implies f^\prime(x) = \frac{1}{2}(x^2f^{\prime\prime}(x - h_1x) - f^{\prime\prime}(x +h_2(1-x))(1-x)^2)$$

$$|f^\prime(x)| \leq \frac{A}{2}(2x^2 -2x+1) < \frac{A}{2} \forall x \in(0,1)$$

share|cite|improve this answer

Here's a hint towards a proof by contradiction. Suppose that there exists $w\in[0,1]$ such that $f'(w) > A/2$. Then the function $f'(x)$ cannot take values very much less than $A/2$ when $x$ is near $w$ - because the derivative of $f'$, namely $f''$, cannot be too large. Can you use the Mean Value Theorem to derive a lower bound on $f'(x)$ that depends on the distance $|x-w|$?

Then notice that we're supposed to have $0 = f(1)-f(0) = \int_0^1 f'(x) \, dx$. A sufficiently large lower bound for $f'(x)$ would contradict this....

share|cite|improve this answer
I didn't manage to solve this, since we haven't gotten to integrals. Is there another way? – Harold Jan 23 '13 at 21:09
Harold: What are the tools you know? – Did Jan 26 '13 at 20:19

Hint :By Rolle's theorem, $\exists c\in[0,1] $ with $f'(c)=0$

Using the Mean value theorem on $f'$ show that this, and the fact $|f''(x)|\leq A $ implies $|f'(x)|\leq\frac{A}{2}$

share|cite|improve this answer
how can you guarantee $|f^\prime(x)| \leq \frac{A}{2}$?(in your last step)) is it always true that $ |x-c| \leq \frac{1}{2}$? – jim Jan 27 '13 at 13:51
One of |c-1| and |c|<=1/2. Should I write a more complete answer? – Ishan Banerjee Jan 27 '13 at 14:07
@IshanBanerjee yes! – draks ... Jan 28 '13 at 20:54

This is not a proof but it is a step in the correct direction.

In order to have the maximum $|f'(x)|$ at some x value you should keep $|f''(x)|$ at the maximum value for the entire range. Therefore $|f''(x)| = A$ and for sake of simplicity we will say $f''(x) = -A$.

Therefore:$f'(x) = C_1-Ax$ for some value $C_1$.

This means:$$f(x) = \int(C_1-Ax)dx = C_2+C_1x-Ax^2/2$$

$$f(0) = C_2+0=0; C_2=0$$ $$f(1) = C_2+C_1-A/2=0; C_1=A/2$$ $$f(x) = Ax/2-Ax^2/2$$ $$f'(x) = A/2-Ax$$

And you should be able to see that the $|f'(x)|$ has maxima at 0 and 1 and both have the value of A/2. This is not a true proof because I have not shown that you cannot arrive at a greater $|f'(x)|$ by using a more complex function of $f''(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.