4

Let $f:[0,1]\longrightarrow\mathbb{R}$ be a function twice differentiable with continous second derivative and $f(1)=f(0)$. The inequality: $$\int_{0}^{1}(f''(x))^2dx\geq 120\left(\int_{0}^{1}xf'(x)dx\right)^2$$ holds?

user137654
  • 285

2 Answers2

4

Let $$ A=\int_0^1xf'(x)\,\mathrm{d}x\tag{1} $$ Since $f(0)=f(1)$, we have $$ \int_0^1f'(x)\,\mathrm{d}x=0\tag{2} $$ $(1)$, $(2)$, and integration by parts gives $$ \begin{align} 2A &=\int_0^1(2x-1)f'(x)\,\mathrm{d}x\\ &=\int_0^1f'(x)\,\mathrm{d}x(x-1)\\ &=\int_0^1x(1-x)f''(x)\,\mathrm{d}x\tag{3} \end{align} $$ Apply Hölder to $(3)$: $$ \begin{align} 4A^2 &\le\int_0^1[x(1-x)]^2\,\mathrm{d}x\int_0^1f''(x)^2\,\mathrm{d}x\\ &=\frac1{30}\int_0^1f''(x)^2\,\mathrm{d}x\tag{4} \end{align} $$ Plugging $(1)$ into $(4)$ yields $$ 120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x\tag{5} $$

Using $f(x)=x(1-x)(1+x(1-x))$, we see that $(5)$ is sharp: both sides equal $\dfrac{24}{5}$.

robjohn
  • 345,667
  • Okay, so using integration by parts and then Holder makes sense but one would first try it on $\int_0^1 x^2 f''(x)dx$. Why did it occur to you that the $f(0)=f(1)$ condition would allow for that $x^2$ to be changed to something else? (Nice answer, btw) – abnry Jun 14 '14 at 04:16
  • 1
    What I wanted was to have $u=x(x-1)$ and $v=f'(x)$ so that the limit terms in the integration by parts would vanish, so I needed $(2x-1),\mathrm{d}x$ instead of $x,\mathrm{d}x$ for $\mathrm{d}u$. I recognized that $f(0)=f(1)$ gives us that $\int_0^1f'(x),\mathrm{d}x=0$ and used it. – robjohn Jun 14 '14 at 04:28
  • Can you, please, provide a solution with random limits of integration, say a and b, instead of 0 and 1. – George R. Mar 14 '17 at 06:23
  • @GeorgeR.: Use $g(u)=f\left(\frac{u-a}{b-a}\right)$ or $f(t)=g(a+(b-a)t)$ and write everything in terms of $u$ instead of $t$. – robjohn Mar 14 '17 at 09:47
0

Since $f(1)=f(0)\Rightarrow \int_0^1f'(x)\,\mathrm{d}x=0$, By integration by parts, \begin{equation}\label{MSE-730580} \int_0^1g(x)f''(x)\mathrm{d}x=g(x)f'(x)\bigg|_0^1-\int_0^1g'(x)f'(x)\mathrm{d}x \tag{1} \end{equation} Let $g(x)$ be a polynomial defined as, $$g(x)=ax^2+bx+c\Rightarrow g'(x)=2ax+b$$ Let $$g(x)f'(x)\bigg|_0^1=0\Rightarrow g(0)=g(1)=0\Rightarrow b=-a$$ we can get, $$g(x)=ax(x-1)\Rightarrow g'(x)=2ax-a \tag{2}$$ by $(1),(2)$, we have $$\int_0^1x(x-1)f''(x)\mathrm{d}x=\int_0^1(1-2x)f'(x)\mathrm{d}x=-2\int_0^1xf'(x)\mathrm{d}x$$ by the Cauchy-Schwarz inequality, \begin{align*} \left(\int_0^1xf'(x)\mathrm{d}x\right)^2 &=\left(\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)f''(x)\mathrm{d}x\right)^2\\ &\leqslant\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x \end{align*} Therefore, $$120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x$$ because, $$\int_0^1\bigg(-\frac{1}{2}x(x-1)\bigg)^2\mathrm{d}x=\frac{1}{120}$$

lingdanda
  • 71
  • 3