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Let $f$ be twice differentiable in $[0,1]$

$f(0) = f(1) = 0$,

$|f''(x)|\le A$.

Prove that $|f'(x)| \le \frac{A}2, \forall x \in [0,1] $.

Well this is what I came up with,

$f'(c_1) = \dfrac{f(x) -f(0)}{x-0} = \dfrac{f(x)}{x}$

$f'(c_2) = \dfrac{f(x) -f(1)}{x-1} = \dfrac{f(x)}{x-1}$

for $0\lt c_1,c_2 \lt 1$

from rolle's theorem we know that there is : $0\lt c_3 \lt 1$

such that

$f'(c_3) =0$

Community
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guynaa
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  • you are correct, i added this to the question – guynaa Jan 12 '14 at 21:39
  • fixed it, hope its ok now – guynaa Jan 12 '14 at 21:41
  • all we say is that the second derivative is bounded by A, for example $f(x) = x \cdot (x-1) \cdot x^5$, why isnt this possible? – guynaa Jan 12 '14 at 21:56
  • First, explain why there's a point $k \in [0, 1]$ with $f'(k) = 0$. From there, we can lead you onward. – John Hughes Jan 12 '14 at 22:00
  • The function you propose certainly has $f(0) = f(1) = 0$. If you set about finding the maximum (absolute) value of the second derivative for that function, and call that number $A$, then you'll find (according to the theorem) that for every $x$ in the interval, you also have $|f'(x)| < A/2$. – John Hughes Jan 12 '14 at 22:02
  • OK; now you've found $k$ (which you call $c_3$) with $f'(k) > 0$. Let's do a proof by contradiction. Suppose that for some point $x_0$, we have $f'(x_0) > A/2$. Let's assume that $x_0 < c_3$, because the other case will be similar. What does the mean value theorem on the interval $[x_0, c_3]$ tell you? It'll tell you something about $f''(c_4)$ for some point $c_4$. But you know something else about $f''(c_4)$, namely, that it's less than $A$. You can use that to conclude something about $|x_0 - c_3|$. – John Hughes Jan 12 '14 at 22:52

2 Answers2

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Proof by contradiction. Suppose there exists a point $x^*$ such that $ f'(x^*) > \frac{A}{2}$. (Note: WLOG, the value is positive.)

From FTC, we get that $ \int _0^1 f'(x) \, dx = f(1) - f(0) = 0$.

Apply the Mean Value Theorem to $f$ on $[0,1]$, then there exists some $c \in [0,1]$ such that $f'(c) = 0 $.

Apply the Mean Value Theorem to $f'$ for the interval between $x^*$ and $c$, since $|f''(x)| \leq A$, we get that $|x^* - c | > \frac{1}{2}$. In particular, if we have 2 values of $c$ such that $ f' (c_1) = f'(c_2) = 0 $ and $ c_1 \leq \frac{1}{2} \leq c_2$, then this leads to a contradiction.

WLOG, $c \leq \frac{1}{2} \leq x^*$. This must hold for all values of $c$ such that $ f'(c) = 0 $. Let $c^*$ be the largest possible value of $c$ such that $f'(c) = 0 $. (Note:$f'(c^*) = 0$ by continuity of $f'$.) This means that for $ x > c^*$, $f'(x) > 0 $.

Since $|f''(x) | < A,$ this tells us that $f'(x)$ cannot decrease too quickly, especially around $ x^*$. Observe that $ \int_{c^*} ^ {1} f'(x) \, dx \geq \int_{x^*- \frac{1}{2}}^{x^*} f'(x) \, dx > \frac{A}{8}.$ (Draw a triangle from the point $\left( x^*, f'(x^*)\right)$.

Similarly, $f'(x)$ cannot increase too quickly, especially around $c^*$. Observe that $\left| \int_0^{c^*} f'(x) \, dx \right| \leq \int_0^{c^*} | f'(x)| \, dx < \frac{A}{8}$. (Draw a triangle from the point $\left(c^*, f'(c^*) = 0 \right)$.

This contradicts the initial fact that $ \int_0^1 f'(x) \, dx = 0 $, since the (potentially) negative portion cannot cancel out the (guaranteed) positive portion.

Calvin Lin
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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.

Feu
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  • Very nice, elementary answer. +1 – DonAntonio Jan 13 '14 at 15:33
  • Nice. I'd be amazed if someone hasn't covered integrals yet, but knows what taylor expansion is. Though your answer was accepted in the other question, also with a condition of no integrals. – Calvin Lin Jan 16 '14 at 02:24
  • @CalvinLin Yeah, I wonder if we can find an even more elementary answer. I find it a little too heavy to be proven by someone taking a first course in calculus. – Feu Jan 16 '14 at 18:46