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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.


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

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

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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....

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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 $|f'(x)|\leq\frac{A}{2}$

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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 $f''(x)$.

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