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This question is related to Daniel Fischer's answer here. Suppose $f$ is a real $C^{1}$ function on $[0, 1]$ such that $f(0) = 0$ and $\int_{0}^{1}f'(x)^{2}\, dx \leq 1$. Then (essentially by Cauchy-Schwarz), we have $\left|\int_{0}^{1}f(x)\, dx\right| \leq 2/3$ as Daniel Fischer stated in his answer.

My question is: Is there an example of a function such that we have equality in $\left|\int_{0}^{1}f(x)\, dx\right| \leq 2/3$? I was thinking of maybe a smoothed version of a function which takes the value $0$ on $[0, 1/2)$ and $4/3$ on $(1/2, 1]$ but that seems complicated to construct, moreover, the derivative in some neighbourhood of $1/2$ might be hard to control.

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I think the absolute values are not needed since we always have $\int_0^1 |f(x)| \, dx \geq \left|\int_0^1f(x)\, dx\right|$ while $f'(x)^2$ is unchanged (except at the cusps). So we can just focus on positive functions. Also, instead of investigating if we can reach $\frac{2}{3}$ now that $1$ was proved to be impossible, I think it would be more interesting to find out the maximum possible value of $\int_0^1 f(x) \, dx$ (and the corresponding function) with the given constraints $f(0) = 0$ and $\int_0^1 f'(x)^2 \, dx \leq 1$. –  Pratyush Sarkar Aug 12 '13 at 3:04

1 Answer 1

up vote 5 down vote accepted

No, we can't ever reach $\frac23$. When applying the Cauchy-Schwarz inequality

$$\int_0^x 1\cdot f'(t)\, dt \leqslant \left(\int_0^x 1^2\,dt\right)^{1/2}\left(\int_0^x f'(t)^2\,dt\right)^{1/2} = \sqrt{x}\left(\int_0^x f'(t)^2\,dt\right)^{1/2},$$

we have two factors that ensure that the inequality $f(x) < \sqrt{x}$ is strict.

  1. For small $x$, the integral $\int_0^x f'(t)^2\,dt$ is small, much smaller than $1$.
  2. We only have equality in the Cauchy-Schwarz inequality if the two functions differ only by a constant factor.

The first defect is minimised if the bulk of the weight of $f'$ is concentrated near $0$, but that increases the difference in the CS inequality.

We can get $\int_0^1 f(x)\,dx = \frac{\sqrt{3}}{3}$ by choosing $f'(x) = \sqrt{3}(1-x)$, and that is the maximal possible value:

$$\begin{align} \int_0^1 f(x)\,dx &= \int_0^1 \int_0^x f'(t)\,dt\,dx\\ &= \iint_{0 \leqslant t \leqslant x\leqslant 1} f'(t)\,dt\,dx\\ &= \int_0^1 \int_t^1 f'(t)\,dx\,dt\\ &= \int_0^1 (1-t)f'(t)\,dt\\ &\leqslant \left(\int_0^1 (1-t)^2\,dt\right)^{1/2}\left(\int_0^1 f'(t)^2\,dt\right)^{1/2}\\ &= \sqrt{\frac13} = \frac{\sqrt{3}}{3}. \end{align}$$

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Woah, clever. I like this solution. –  ADF Aug 12 '13 at 15:05

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