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Let $f\in C^{1}[0,1]$ such that $f(0)=f(1)=0$. Show that $$\left(\int_{0}^{1}f(x)dx\right)^2\le\dfrac{1}{12}\int_{0}^{1}|f'(x)|^2dx.$$

I think we must use Cauchy-Schwarz inequality $$\int_{0}^{1}|f'(x)|^2dx\ge \left(\int_{0}^{1}f(x)dx\right)^2$$ but this maybe is not useful to problem.

The coefficient $\dfrac{1}{12}$ is strange. How find it ?

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  • $\begingroup$ In the CS inequality won't it be $|f(x)|^2$ in the LHS? $\endgroup$ – Samrat Mukhopadhyay Nov 12 '14 at 15:47
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    $\begingroup$ The answer by Joel is wrong. See my solution. $\endgroup$ – xpaul Nov 13 '14 at 20:54
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Using integration by parts and $f(0)=f(1)=0$, we have: $$ \int_0^y f(x)dx = \int_0^y(y-x)f'(x) dx$$ and $$ \int_y^1 f(x)dx = \int_y^1(y-x)f'(x) dx$$ for all $y\in [0,1]$.

Taking $y=1/2$ and using Cauchy-Schwarz inequality we get: $$ \left(\int_0^{1/2} f(x)dx\right)^2 = \left(\int_0^{1/2}({1/2}-x)f'(x) dx\right)^2 $$ $$\leq \int_0^{1/2}({1/2}-x)^2 dx \int_0^{1/2}(f'(x))^2 dx = \frac{1}{24} \int_0^{1/2}(f'(x))^2 dx$$ and $$ \left(\int_{1/2}^1 f(x)dx\right)^2 = \left(\int_{1/2}^1({1/2}-x)f'(x) dx\right)^2 $$ $$\leq \int_{1/2}^1({1/2}-x)^2 dx \int_{1/2}^1(f'(x))^2 dx = \frac{1}{24} \int_{1/2}^1(f'(x))^2 dx.$$ So:

$$\frac{1}{2} \left(\int_0^{1} f(x)dx\right)^2\leq \left(\int_0^{1/2} f(x)dx\right)^2 + \left(\int_{1/2}^1 f(x)dx\right)^2 $$ $$\leq \frac{1}{24} \int_0^{1/2}(f'(x))^2 dx+\frac{1}{24} \int_{1/2}^1(f'(x))^2 dx = \frac{1}{24} \int_0^{1}(f'(x))^2 dx.$$

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In fact $$ \left(\int_0^1(2x-1)f'(x)dx\right)^2\le\int_0^1(2x-1)^2dx\int_0^1(f'(x))^2dx. \tag1 $$ But $$ \int_0^1(2x-1)f'(x)dx =-2\int_0^1f(x)dx$$ by using the Integration-by-Parts formula and $$ \int_0^1(2x-1)^2dx=\frac{1}{3}. $$ Putting these two into (1), you can get the desired inequality.

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Using integration by parts we find that $$\int_0^1 f(x)dx = - \int_0^1 xf'(x)dx$$

Then Cauchy-Schwarz will give $$\left( \int_0^1 x f'(x) dx \right)^2 \le \left( \int_0^1 |x f'(x)| dx \right)^2 \le \int_0^1 x^2 dx \cdot \int_0^1 |f'(x)|^2 dx$$

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  • $\begingroup$ Note we use the condition $f(0)=f(1)=0$ when we performed integration by parts. $\endgroup$ – Joel Nov 12 '14 at 15:49
  • $\begingroup$ After seeing the two downvotes to this question, I realized that this method gives the wrong constant. When you perform the integration on $\int_0^1 x^2 dx$ the result is 1/3 and not 1/12. $\endgroup$ – Joel Nov 12 '14 at 20:15
  • $\begingroup$ @xpaul For a day, this was the only answer. New answers did not come about until an hour ago. I didn't claim it settled the matter, but it does give a constant independent of $f$ for the inequality. It's a basic first step for attacking the problem. $\endgroup$ – Joel Nov 13 '14 at 21:40
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The parameter $\frac{1}{12}$ is kind of weird, it has the factor of 3, when the problem is about integration or derivation,we can assume it's about quadratic function.

what's more, as $f(0) = f(1) = 0$, we assume the equality stands when the function is symmetrical.

after trying, we can see function $f(x) = x(1-x)$ can satisfy the equality, while $f'(x) = -2x+1 = -2(x-\frac{1}{2})$.

in the light of the fact above, we substitute $\frac{1}{12}$ with $\int_{0}^{1}|x-\frac{1}{2}|dx$, using C-S inequality , we have:

$$ \frac{1}{12}\int_{0}^{1}|f'(x)|^2dx = \int_{0}^{1}|x-\frac{1}{2}|^2dx\int_{0}^{1}|f'(x)|^2dx \geq (\int_{0}^{1}|f'(x)(x-\frac{1}{2})|dx)^2 $$

$$ \int_{0}^{1}|f'(x)(x-\frac{1}{2})|dx \geq |\int_{0}^{1}f'(x)(x-\frac{1}{2})dx| = |\int_{0}^{1}f(x)dx| $$

so,

$$ (\int_{0}^{1}f(x)dx)^2 \leq \frac{1}{12}\int_{0}^{1}|f'(x)|^2dx $$

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