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Here's the problem:

Suppose the set $\{x\mid f(x)\not = 0,x\in[a,b]\}$ is not empty, and $f$ is differentiable on $[a,b]$, with $f(a)=f(b)=0$.

Prove that $\exists c$, such that $$|f'(c)|>\frac{4}{(b-a)^2}\int_{a}^{b}|f(x)|dx$$

My attempt is that first, to prove that exists $c$ s.t. $$|f'(c)|>\frac{1}{b-a}\int_{a}^{b}|f'(x)|dx$$ and then, use the positive variation to prove that $$\int_{a}^{b}|f'(x)|dx\ge\frac{2}{(b-a)}\int_{a}^{b}|f(x)|dx$$ since $|f(x)|\le P_F(b)$, with $P_F(x)$ is positive variation function of $f$.

But the coefficient is only $\frac{2}{(b-a)^2}$, not $\frac{4}{(b-a)^2}$, which bother me a lot.

Both coefficient in my two inequality cannot be improved (as far as I know..) since the former one can be approximated by letting $f$ be a tent function and the latter one can be approximated by letting $f$ be a constant greater than zero.(Of course both example should be mollified so that $f$ can be differentiated.)

Hope to find some valid method, thanks for your attention!

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Golbez
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  • Your attempt isn't right, since $f'$ needn't be integrable. – Yai0Phah Sep 07 '13 at 03:20
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    WLOG, suppose that $a=-1$ and $b=1$, and $M$ is the upper-bound of $\lvert f'(x)\rvert$, we can see that $(x,\lvert f(x)\rvert)$ lies in the triangle $(-1,0),(1,0),(0,M)$, so $\int_{-1}^1\lvert f(x)\rvert dx\le M$ – Yai0Phah Sep 07 '13 at 03:22
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    For the strict inequality, let $M=\sup\lvert f'(x)\rvert$, show that $M>0$, and $\int_{-1}^1\lvert f(x)\rvert dx<M$ and then take an appropriate $c$. – Yai0Phah Sep 07 '13 at 03:29
  • @FrankScience The geometric meaning now seems obvious, but could you write it in detail (the graph lies in a triangle )and make it an answer so that I can accept it? Thanks! – Golbez Sep 07 '13 at 03:41
  • It's a chance for you to write it on your own. You could accept your own answer. For now I have no interest in writing up a detailed answer (I'm working on another problem). – Yai0Phah Sep 07 '13 at 03:48
  • @FrankScience All right. Last question, can $f'$ possibly not be integrable? Since $f$ is differentiable in a closed interval, condition in fundamental theorem of calculus is satisfied. – Golbez Sep 07 '13 at 04:07
  • @FrankScience Sorry, seems that I didn't mention $f(x)$ is differentiable on $[a,b]$. – Golbez Sep 07 '13 at 04:09
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    Barrow's formula needs the integrability of $f'$. There are examples that the discontinuities of $f'$ could be very general. It's a very advanced topic, see here or here. – Yai0Phah Sep 07 '13 at 04:14

1 Answers1

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With the help of @FrankScience , here's my new attempt.

WLOG, let $a=-1,b=1$, the inequality becomes $$|f'(c)|>\int_{-1}^1|f(x)|dx$$. Here we must prove that the graph $(x,|f(x)|)$ lies in the triangle made from $(-1,0),(1,0),(0,M)$. Where $M=\sup|f'(x)|$.

That is ,we need $x\in[-1,0]\Rightarrow|f(x)|\le M(x+1)$ and $x\in[0,1]\Rightarrow|f(x)|\le -M(x-1)$.

For the first inequality, note that $$|f(x)|=\left|\int_{-1}^x f'(s)ds\right|\le M(x+1)$$ Second inequality holds since $$|f(x)|=\left|\int_{x}^{1} f'(s)ds\right|\le M(1-x)$$ That is, $$\int_{-1}^{1}|f(x)|dx\le \sup |f'(x)|$$ To get the equality, $f(x)$ should be equal to $M(x+1)$ as well as $M(1-x)$ almost every where respectively on these two intervals, which cannot be satisfied since $f$ is differentiable. So strict inequality holds

Golbez
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