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Let $f$ be a continuously differentiable real valued function on $[0,1]$.

It is given that $\displaystyle \int_{\frac{1}{3}}^{\frac{2}{3}}f(x) dx=0$

Find the minimum value of $\dfrac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_{0}^{1} f(x) dx \right)^2}$

I tried to use Cauchy-Schwarz to show that $$\frac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_0^1 f(x) dx \right)^2} \ge \frac{\left( \int_0^1 \bigl| f(x)f'(x) \bigr| dx \right)^2}{ \left( \int_0^1 f(x) dx \right)^2} \ge \frac{ f(1)^2 - f(0)^2}{2 \left( \int_0^1 f^2 (x) dx \right)^2}$$ But I can't proceed from here.

Also, I don't know how to use the condition $\int_{1/3}^{2/3}f(x) dx=0$

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  • $\begingroup$ Hope you don't mind my edit, and please feel free to overwrite. Is there a typo in the last denominator? What is $f^2(x)$ supposed to mean? The second derivative? If so then it should be $f^{(2)}(x)$. $\endgroup$ – Lee David Chung Lin Nov 5 '18 at 16:50
  • $\begingroup$ No, it means the square of $f(x)$ $\endgroup$ – Legend Killer Nov 5 '18 at 17:25
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The minimum ratio is $27$, achieved by a piecewise quadratic polynomial in $x$.

This is sort of expected. When one throw the functional

$$\int_0^1 f'(x)^2 dx$$ to Euler-Lagrange equation and subject it to the constraints $$\int_0^1 f(x) dx = \text{constant}\quad\text{ and }\quad \int_{1/3}^{2/3} f(x) dx = 0$$ One find $f''(x)$ has to be a piecewise constant. It takes one value over $[0,\frac13) \cup (\frac23,1]$ and another value over $(\frac13,\frac23)$. What I has done is use a CAS to pin down the correct piecewise quadratic polynomial and then verify it give us the minimum.


Let $X = \mathcal{C}^1[0,1]$ and $P,Q,C : X \to \mathbb{R}$ be the functionals over $X$ defined by

$$P(f) = \int_0^1 f'(x)^2 dx,\quad Q(f) = \int_0^1 f(x) dx,\quad\text{ and }\quad C(f) = \int_{1/3}^{2/3} f(x) dx$$

The question can be rephrased as

What is the minimum of the ratio $\frac{P(f)}{Q(f)^2}$ for $f \in X$ subject to the constraint $C(f) = 0$.

Since the ratio and constraint are both invariant under scaling of $f$ by constant, we can restrict our attention to those $f$ which satisfies $C(f) = 0$ and $Q(f)$ equal to a specific constant.

For any $K \in \mathbb{R}$, let $Y_K = \big\{\; f \in X : C(f) = 0, Q(f) = K\; \big\}$.

Consider following function over $[0,1]$

$$g(x) = \begin{cases} 4 - 27x^2, & x \in [0,\frac13]\\ 54(x^2-x) + 13, & x \in [\frac13,\frac23]\\ 4 - 27(1-x)^2, & x \in [\frac23,1] \end{cases}$$ We have $$ g'(x) = \begin{cases} -54x, & x \in [0,\frac13]\\ 54(2x-1),& x \in [\frac13,\frac23]\\ 54(1-x),& x \in [\frac23,1] \end{cases},\quad g''(x) = \begin{cases} -54, & x \in [0,\frac13) \cup (\frac23,1]\\ 108, & x \in (\frac13,\frac23) \end{cases} $$ It is not hard to see $g \in X$. With a little bit of effort, one can verify $P(g) = 108$, $Q(g) = 2$ and $C(g) = 0$. This means $g \in Y_2$.

For other $f \in Y_2$, it is easy to see $\eta = f - g \in Y_0$. We can decompose $P(f)$ as follows

$$P(f) = \int_0^1 (g'(x)+\eta'(x))^2 dx = \int_0^1 (g'(x)^2 + \eta'(x)^2 + 2g'(x)\eta'(x)) dx$$ Let us look at the cross term. Integrate by part and using the fact $g'(0) = g'(1) = 0$, we find

$$\begin{align}\int_0^1 g'(x)\eta'(x) dx &= [ g'(x) \eta(x) ]_0^1 - \int_0^1 g''(x)\eta(x) dx\\ &= 54 \int_0^1 \eta(x) dx - 162\int_{1/3}^{2/3}\eta(x)dx\\ &= 54 Q(\eta) - 162 C(\eta) \end{align} $$ Since $\eta \in Y_0$, $Q(\eta) = C(\eta) = 0$ and the cross term goes away. As a result,

$$P(f) = P(g) + P(\eta) \ge P(g)$$ because $P(\eta)$ is non-negative. Together with $Q(f) = Q(g) = 2$, we get

$$\frac{P(f)}{Q(f)^2} \ge \frac{P(g)}{Q(g)^2} = \frac{108}{2^2} = 27$$

As a result,

$$\min\left\{ \frac{P(f)}{Q(f)^2} : f \in X, C(f) = 0 \right\} = \min\left\{ \frac{P(f)}{Q(f)^2} : f \in Y_2 \right\} = 27$$

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  • $\begingroup$ Did the Lagrange Euler equation force you to the fact that $f''(x)$ is a constant?Moreover $f$ may not be double differentiable $\endgroup$ – Legend Killer Nov 6 '18 at 2:40
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    $\begingroup$ @LegendKiller within the framework, it does force local extremum of the ratio to has piecewise constant $f''$. However, there are loopholes, e.g. the actual $f$ may not be doubly differentiable or what one get is only a local extremum. That's why after EL tell you what the minimum "should" be, one need to verify it is indeed the correct answer. $\endgroup$ – achille hui Nov 6 '18 at 3:43

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