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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

2 Answers 2

up vote 7 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

Since another question was linked here as duplicate I would like to give an alternative solution to the problem $$ \max\int_0^1f(x)\,dx,\qquad f(0)=0,\ \int_0^1\dot f^2(x)\,dx\le 1. $$ If we, for example, did not get the intuition for the artistic solution by CS above, we can try to approach the problem in a more standard way via the variational calculus.

  1. Formalization: we have the minimization problem with one functional constraint and one fixed point (the left end) $$ \min\int_0^1(-f(x))\,dx,\qquad f(0)=0,\ \int_0^1\dot f^2(x)\,dx\le 1. $$
  2. Use Lagrange multiplier method to include the constraint into the optimization with $\lambda\ge 0$ $$ \min\int_0^1(-f(x)+\lambda(\dot f^2(x)-1))\,dx=\min\int_0^1\bigl(-f(x)+\lambda\dot f^2(x)\bigr)\,dx-\lambda,\qquad f(0)=0 $$
  3. Write down the Euler-Lagrange equation $$ -1+2\lambda\ddot f(x)=0,\quad f(0)=0. $$ Here we see that $\lambda\ne 0$ (otherwise no solutions), so $\lambda>0$. [It explains, in particular, why the constraint must be active, i.e. $\|\dot f\|_2^2=1$, because the complementary slackness principle tells us that $\lambda(\|\dot f\|_2^2-1)=0$.]
  4. Add the natural boundary condition at the free right end $$ 2\lambda\dot f(1)=0. $$
  5. The solution to the E.-L. equation with two boundary conditions gives $$ f(x)=\frac{1}{4\lambda}x(x-2). $$
  6. Using the fact that the constraint must be active, we find $\lambda$ from the equation $$ 1=\int_0^1\dot f^2(x)\,dx=\int_0^1\frac{1}{4\lambda^2}(x-1)^2\,dx\quad\Rightarrow\quad\lambda=\frac{1}{2\sqrt{3}}. $$ Thus $$ f(x)=\frac{\sqrt{3}}{2}x(x-2). $$
  7. Convexity of the problem makes the Euler-Lagrange equation sufficient for the minimum (the same proof as in Theorem 4.3, page 12).
  8. Change the sign for $f$ to get the solution to the original problem.
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I fail to see how to show convexity of S and then convexity of $L(x,f(x),f'(x))=-f(x)+\lambda f'(x)^2$, this is needed in step 7. Could you give me a hint on that? – Cyclone Aug 31 at 16:20
The set $S$ is convex as an intersection of the linear subspace (hyperplane $f(0)=0$) and the ball $\|f'\|_2\le 1$ (convex by the triangle inequality for $L_2$ norm). The Lagrangian is convex since the function $f(u,v)=-u+\lambda v^2$ is convex for $\lambda\ge 0$ (the sum of linear $-u$ and convex $\lambda v^2$ is convex). – A.G. Aug 31 at 16:31

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