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Question:

Let $f:[a,b] \to \Bbb R$ be a continuously differentiable function s.t $f(a)=f(b)=0$ Prove that exists a point $c \in (a,b)$ such that $$ |f'(c)| \ge \frac 4{(b-a)^2} \int ^b_a f(x) dx $$ What we did: We tried using the integral mean value theorem and then using LaGrange theorem, but ended with a sum of 2 points $(f'(y_1) + f'(y_2))$ in $[a,b]$ that we didn't know how to connect to $|f'(c)|$.

2nd try (using Newton Leibnitz, and then LaGrange theorem twice): $|\frac 4{(b-a)^2}\int_a^b f(x)dx| = |\frac 4{(b-a)^2}(F(a)-F(b))|= |\frac 4{(b-a)} \frac{(F(a)-F(b))}{b-a}| = |\frac 4{(b-a)} f(c_1)|= |\frac 4{(b-a)} \frac {(f(c_1)-f(a)) (c_1-a)}{c_1-a}|= |\frac {4(c_1-a)}{(b-a)} \frac {(f(c_1)-f(a))}{c_1-a}|= |\frac {4(c_1-a)}{(b-a)} f'(c)|$

our problem with this try is that we don't know how we can bound the last expression with f'(c)

jreing
  • 3,297
  • Is $f$ nonnegative? – detnvvp Oct 15 '13 at 19:33
  • It doesn't matter if $f$ is nonnegative, because the integrals where $f$ is positive and negative can cancel out somewhat, but the maximal absolute value of the derivative will stay the same. So you might as well assume $f$ is nonnegative. – user2566092 Oct 15 '13 at 19:41
  • f' is continuous, so you can use intermediate value theorem. – achille hui Oct 15 '13 at 19:42
  • Do you mean we should use the (regular) intermediate value theorem on f' or on f? Could you please elaborate a little... thx @achille hui – jreing Oct 17 '13 at 08:47
  • $f$ continuous differentiable means $f'$ continuous. Apply IVT to $f'$, you will find $f'(y_1) + f'(y_2) = 2 f'(\xi)$ for some $\xi$ between $y_1$ and $y_2$. – achille hui Oct 17 '13 at 09:04
  • I posted a new try with a different direction now... I'll try checking out your suggestion now – jreing Oct 17 '13 at 09:09

1 Answers1

1

Let $F$ be an antiderivative (exist by FTC) of $f$. $F$ is twice differentiable by the given conditions.

By Taylor's theorem, there exists $c_1\in (a,b)$ and $c_2\in (a,b)$ such that $$F(b)=F(a)+(b-a)F'(a)+\frac{(b-a)^2}{2}F''(c_1)$$ and $$F(a)=F(b)+(a-b)F'(b)+\frac{(a-b)^2}{2}F''(c_2)$$

Subtract the equations using $F'(a)=F'(b)$, we have $2(F(b)-F(a))=\frac{(a-b)^2}{2}(F''(c_2)-F''(c_1))=\frac{(a-b)^2}{2}F''(c)$, last equality follows from intermediate value property of derivative(doesn't even need to be continuous!). Therefore $|f'(c)|\geq f'(c)=\frac{4(F(b)-F(a))}{(b-a)^2}=\frac{4}{(b-a)^2}\int_a^b f(x)dx$