Minimum value of the function over all non zero continuously differentiable functions

Question

Find the minimum value of $$ \frac{\int_{0}^{1} x^{2}\left(f^{\prime}(x)\right)^{2} d x}{\int_{0}^{1} x^{2}(f(x))^{2} d x} $$ over all nonzero continuously differentiable functions $f:[0,1] \rightarrow \mathbb{R}$ with $f(1)=0$. My approach was letting $$\int_{0}^{x} f'(x) dx = f(x)-f(0) $$ I tried substituiting, but it wasnt helping at all. (Variational calculus based solution also works other than integral bounding.) Does maximimizing denominator helps ?

Answer 1

Just for brevity, I'll define operators $A: C^1([0,1]) \to \mathbb{R}$ and $Q: \{ f \in C^1([0,1]) : f(1)=0 \} \setminus \{x \mapsto 0\} \to \mathbb{R}$:

$$ \begin{align*} A(f) &= \int_0^1 x^2 f(x)\, dx \\ Q(f) &= \frac{A((f')^2)}{A(f^2)} \end{align*} $$

Above and throughout, I'm using notation like $f^2$ to always mean the product $f \cdot f$, not the composition $f \circ f$.

Suppose function $f$ minimizes $Q$.

First, we have $A(f^2) > 0$ and $A((f')^2) > 0$. (The zero function is specifically excluded from the domain, and $f$ is not a constant since $f(1)=0$, so $f'$ is not the zero function either.)

Take any function $g \in C^1([0,1])$ with $g(1)=0$, and any real number $\delta$ (except the $\delta$, if any, which would cause $f+\delta g$ to be the zero function). Since $A$ is linear,

$$ \begin{align*} \frac{d}{d\delta} Q(f+\delta g) &= \frac{d}{d\delta}\left[ \frac{A\big((f' + \delta g')^2)}{A\big((f+\delta g)^2\big)} \right] \\ &= \frac{2 A\big((f'+\delta g') g'\big) A\big((f+\delta g)^2\big) - 2A\big((f'+\delta g')^2\big) A\big((f+\delta g)g\big)}{\left[A\big((f+\delta g)^2\big)\right]^2} \end{align*} $$

For $f$ to be a minimum, the derivative above must equal $0$ when $\delta=0$, giving

$$ A(f'g')A(f^2) - A((f')^2) A(fg) = 0 $$

$$ A(f'g') - Q(f) A(fg) = 0 \tag{1} $$

This must be true of any $g$ satisfying the criteria; consider if $g$ is the function $T_{\alpha,x_0}$ defined in terms of real numbers $\alpha > 0$ and $x_0 \in (0,1)$:

$$ T_{\alpha,x_0}(x) = \begin{cases} -1 & x < x_0-\alpha \\ -1 + \frac{1}{2\alpha^2} (x-x_0+\alpha)^2 & x_0-\alpha \leq x < x_0 \\ - \frac{1}{2\alpha^2} (x-x_0-\alpha)^2 & x_0 \leq x \leq x_0 + \alpha \\ 0 & x > x_0 + \alpha \end{cases} $$

$T_{\alpha,x_0}$ is designed so its derivative has a triangle shape:

$$ T_{\alpha,x_0}'(x) = \begin{cases} 0 & x < x_0-\alpha \\ \frac{1}{\alpha^2}(x-x_0+\alpha) & x_0-\alpha \leq x < x_0 \\ \frac{1}{\alpha^2}(x_0+\alpha-x) & x_0 \leq x \leq x_0+\alpha \\ 0 & x > x_0+\alpha \end{cases} $$

Since $f$ is continuous on $[0,1]$, it is bounded on $[0,1]$, and so

$$ \lim_{\alpha \to 0^{+}} A(f T_{\alpha,x_0}) = - \int_0^{x_0} x^2 f(x)\, dx $$

By continuity of $f'$,

$$ \lim_{\alpha \to 0^{+}} A(f' T_{\alpha,x_0}') = x_0^2 f'(x_0) $$

Plugging $g=T_{\alpha,x_0}$ into equation (1) and taking the limit,

$$ 0 = \lim_{\alpha \to 0^{+}} \big[ A(f' T_{\alpha,x_0}') - Q(f) A(f T_{\alpha,x_0}) \big] = x_0^2 f'(x_0) + Q(f) \int_0^{x_0} x^2 f(x)\, dx $$

Since this is true for every $x_0 \in (0,1)$, and since it can be solved for $f'(x_0)$ giving a formula with defined derivative, this proves the minimal $f$ is in fact twice differentiable, satisfying:

$$ x_0^2 f''(x_0) + 2 x_0 f'(x_0) + Q(f) x_0^2 f(x_0) = 0 $$

(If we knew ahead of time $f$ had a second derivative, doing an integration by parts on $A(f'g')$ would get the same differential equation more easily. I'm not sure if there's a simpler way or general result which shows $f''$ exists here.)

We can now rename $x_0$ as $x$ without as much confusion, and we're looking for solutions to the differential equation

$$ x f''(x) + 2 f'(x) + q x f(x) = 0 $$

where $q = Q(f)$ is for now just a positive constant. If $h(x) = x f(x)$, then $h'(x) = x f'(x) + f(x)$ and $h''(x) = x f''(x) + 2 f'(x)$, so

$$ h''(x) + q h(x) = 0 $$

$$ h(x) = c_1 \sin(\sqrt{q} x + c_2) $$

$$ f(x) = \frac{c_1 \sin(\sqrt{q} x + c_2)}{x} $$

Since $f$ has a finite limit $\lim_{x \to 0^{+}} f(x) = f(0)$, $c_2$ must be a multiple of $\pi$. Since $\sin(\sqrt{q} x + m \pi) = (-1)^m \sin(\sqrt{q} x)$ and $c_1$ already can change sign, take $c_2=0$ without loss of generality.

Since $f(1)=0$, $\sqrt{q}$ must be a multiple of $\pi$, so say $\sqrt{q} = n \pi$.

Multiplying $f$ by a constant doesn't change $Q(f)$, since

$$ Q(C f) = \frac{A((Cf')^2)}{A((Cf)^2)} = \frac{C^2 A((f')^2)}{C^2 A(f^2)} = \frac{A((f')^2)}{A(f^2)} = Q(f) $$

So we can assume $c_1=1$, knowing that for any solution $f$, $c_1 f$ is another solution.

This leaves the formula

$$ f(x) = \frac{\sin(n \pi x)}{x} $$

with $n \in mathbb{Z}$. Then

$$ A(f^2) = \int_0^1 \frac{x^2 \sin^2(n \pi x)}{x^2} dx = \int_0^1 \frac{1}{2}(1-\cos(2 n \pi x))\, dx = \frac{1}{2} $$

$$ \begin{align*} f'(x) &= \frac{n \pi \cos(n \pi x)}{x} - \frac{\sin(n \pi x)}{x^2} \\ (f'(x))^2 &= n^2 \pi^2 \frac{\cos^2 (n \pi x)}{x^2} - 2 n \pi \frac{\cos(n \pi x) \sin(n \pi x)}{x^3} + \frac{\sin^2(n \pi x)}{x^4} \\ A((f')^2) &= \int_0^1 \left[ n^2 \pi^2 \cos^2(n \pi x) - 2 n \pi \frac{\cos(n \pi x) \sin(n \pi x)}{x} + \frac{\sin^2(n \pi x)}{x^2} \right]\, dx \\ A((f')^2) &= \int_0^1 \left[ \frac{n^2 \pi^2}{2} (1+\cos(2n \pi x)) - n \pi \frac{\sin(2n \pi x)}{x} + \frac{1-\cos(2 n \pi x)}{2x^2}\right]\, dx \\ A((f')^2) &= \frac{n^2 \pi^2}{2} + \lim_{a \to 0^{+}} \left[ - \frac{1-\cos(2n \pi x)}{2x} \right]_{x=a}^{x=1} + \int_a^1 \left[ - n \pi \frac{\sin(2n \pi x)}{x} + 2n \pi \frac{\sin(2n \pi x)}{2x}\right]\, dx \\ A((f')^2) &= \frac{n^2 \pi^2}{2} \end{align*} $$

Finally, $Q(f) = n^2 \pi^2$. Since this matches the choice $n \pi = \sqrt{q} = \sqrt{Q(f)}$ in the differential equation, this $f$ does satisfy the differential equation for every non-zero integer $n$ and is a potential local minimum. $n$ can't be zero since that would make $f$ the zero function. So the global minimum is with $n = \pm 1$, giving minimum value $\pi^2$, and the only functions which attain that minimum are $$f(x) = C \frac{\sin(\pi x)}{x}$$ for any real non-zero $C$.