Given $\int_{\frac13}^{\frac23}f(x)dx=0$, how to prove $4860(\int_0^1f(x)dx)^2\le 11\int_0^1|f''(x)|^2dx$?

Solution 1:

This is essentially following the steps in my answer to a quasi-similar question. I'm not going to explain how I find the function $g(x)$ below.

Let $X = \mathcal{C}^2[0,1]$ and $P,Q,C : X \to \mathbb{R}$ be 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

Given $f \in X$ with $C(f) = 0$, how to verify $\;P(f) \ge \frac{4860}{11} Q(f)^2$?

Since both the inequality and constraint is homogeneous in scaling of $f$ by a constant. We can restrict our attention to those $f$ which satisfies $C(f) = 0$ and $Q(f) = 1$.

Consider following functions

$$\phi(x) = x^4 - \frac12 x^2 + \frac{29}{6480} \quad\text{ and }\quad \psi(x) = \begin{cases} \left(\frac13-x\right)^4, & x \le \frac13\\ 0, & \frac13 \le x \le \frac23\\ \left(x - \frac23\right)^4, & x \ge \frac23 \end{cases} $$ Combine them and define another function $g(x)$ by $$g(x) = -\frac{405}{11}\left[ \phi\left(x-\frac12\right) - \frac32 \psi(x) \right]$$ It is not hard to check

1. $g(x) \in \mathcal{C}^3[0,1] \subset X$.
2. $C(g) = 0$, $Q(g) = 1$.
3. $g''(0) = g'''(0) = g''(1) = g'''(1) = 0$
4. $g''''(x) = \frac{4860}{11}$ for $x \in [0,\frac13)\cup (\frac23,1]$
5. $g''''(x) = -\frac{9720}{11}$ for $x \in (\frac13,\frac23)$
6. $P(g) = \frac{4860}{11}$.

For any $f \in X$ with $C(f) = 0, Q(f) = 1$, let $\eta = f - g$, we have

$$\begin{align} & P(f) - P(g) - P(\eta)\\ = & 2\int_0^1 g''(x)\eta''(x) dx\\ = & 2\int_0^1 ( g''(x)\eta'(x))' - g'''(x)\eta'(x) dx\\ = & 2\int_0^1 ( g''(x)\eta'(x) - g'''(x)\eta(x))' + g''''(x)\eta(x)dx\\ = &2\left\{\left[ g''(x)\eta'(x) - g'''(x)\eta(x) \right]_0^1 + \frac{4860}{11}(Q(\eta)-C(\eta)) -\frac{9720}{11}C(\eta)\right\} \end{align} $$ What's in the square bracket vanish because of $(3)$. The remain terms vanish because

• $Q(\eta) = Q(f) - Q(g) = 1 - 1 = 0$.
• $C(\eta) = C(f) - C(g) = 0 - 0 = 0$.

Together with the fact $P(\eta)$ is non-negative, we obtain:

$$P(f) = P(g) + P(\eta) \ge P(g) = \frac{4860}{11}$$.

Solution 2:

I have worked it out by myself! Here's my answer. In order to vanish the values of $f$,$f'$at $0$,$1$,we do \begin{align*} \int_0^{\frac13}x^2f''(x)d\,x =&\int_0^{\frac13}x^2d\,f'(x)=x^2f'(x)\big|_0^{\frac13}-\int_0^{\frac13}f'(x)\cdot 2x d\,x\\ =&x^2f'(x)\big|_0^{\frac13}-2xf(x)\big|_0^{\frac13}+2\int_0^{\frac13}f(x)d\,x\\ =&\frac19f'(\frac13)-\frac23f(\frac13)+2\int_0^{\frac13}f(x)d\,x\triangleq r\tag{1}\\ \int_{\frac23}^1(x-1)^2f''(x)d\,x=&\int_{\frac23}^1(x-1)^2d\,f'(x)\\ =&(x-1)^2f'(x)\big|_{\frac23}^1-\int_{\frac23}^1f'(x)\cdot 2(x-1)d\,x\\ =&(x-1)^2f'(x)\big|_{\frac23}^1-2(x-1)f(x)\big|_{\frac23}^1+2\int_{\frac23}^1f(x)d\,x\\ =&-\frac19f'(\frac23)-\frac23f(\frac23)+2\int_{\frac23}^1f(x)d\,x\triangleq s\tag{2} \end{align*} Take $a,b$ to be confirmed,we have \begin{align*} &\int_{\frac13}^{\frac23}(x^2+ax+b)f''(x)d\,x \\ =&f'(x)(x^2+ax+b)\big|_{\frac13}^{\frac23}-f(x)(2x+a)\big|_{\frac13}^{\frac23}+2\int_{\frac13}^{\frac23}f(x)d\,x \\ =&f'(\frac23)(\frac49+\frac23a+b)-f'(\frac13)(\frac19+\frac13a+b)-f(\frac23)(\frac43+a)+f(\frac13)(\frac23+a) \end{align*} By comparing the ratio of coefficients of $f(\frac13)$,$f(\frac23)$,$f'(\frac13)$and$f'(\frac23)$in $(1)$,$(2)$,we pick $a=-1$,$b=\frac16$,and get \begin{align*} \int_{\frac13}^{\frac23}(2x^2-2x+\frac13)f''(x)d\,x=-\frac19f'(\frac23)+\frac19f'(\frac13)-\frac23f(\frac23)-\frac23f(\frac13) \triangleq -t\tag{3} \end{align*} Associate the above three formulae and use Cauchy-Schwarz inequality,we get \begin{align*} \Big(\int_0^{\frac13}x^4d\,x \Big)\Big(\int_0^{\frac13}(f''(x))^2d\,x\Big)\geqslant &r^2\\ \Big(\int_{\frac23}^1(x-1)^4d\,x\Big)\Big(\int_{\frac23}^1(f''(x))^2d\,x\Big)\geqslant & s^2\\ \Big(\int_{\frac13}^{\frac23}(2x^2-2x+\frac13)^2d\,x\Big)\Big(\int_{\frac13}^{\frac23}(f''(x))^2d\,x\Big)\geqslant & t^2 \end{align*} that is \begin{align*} \frac{1}{1215}\int_0^{\frac13}(f''(x))^2d\,x\geqslant & r^2\\ \frac{1}{1215}\int_{\frac23}^1(f''(x))^2d\,x\geqslant &s^2\\ \frac1{1215}\int_{\frac13}^{\frac23}(f''(x))^2d\,x\geqslant & \frac19t^2 \end{align*} Sum up the above three formulae,we can get \begin{align*} \frac1{1215}\int_0^1(f''(x))^2d\,x\geqslant r^2+s^2+\frac19t^2\tag{4} \end{align*} By using Cauchy's ineuality again,we have \begin{align*} (r+s+t)^2=(1\cdot r+1\cdot s+3\cdot \frac13t)\leqslant 11 (r^2+s^2+\frac19t^2)\tag{5} \end{align*} Finally we get \begin{align*} \frac1{1215}\int_0^1(f''(x))^2d\,x\geqslant &\frac1{11}(r+s+t)^2=\frac{1}{11}\Big(2\int_0^{\frac13}f(x)d\,x+2\int_{\frac23}^1f(x)d\,x \Big)^2\\ =&\frac4{11}\Big(\int_0^1f(x)d\,x \Big)^2 \end{align*}

Solution 3:

By integration by parts, we have $$ \begin{aligned} \int_0^{\frac{1}{3}}g_1(x)f''(x)\mathrm{d}x &=g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}-\int_0^{\frac{1}{3}}g_1'(x)f'(x)\mathrm{d}x\\ &=g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}-g_1'(x)f(x)\bigg|_0^{\frac{1}{3}}+\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x \end{aligned}\tag{1} $$ $$ \begin{aligned} \int_{\frac{1}{3}}^{\frac{2}{3}}g_2(x)f''(x)\mathrm{d}x &=g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}-\int_{\frac{1}{3}}^{\frac{2}{3}}g_2'(x)f'(x)\mathrm{d}x\\ &=g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}-g_2'(x)f(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}+\int_{\frac{1}{3}}^{\frac{2}{3}}g_2''(x)f(x)\mathrm{d}x \end{aligned}\tag{2} $$ $$ \begin{aligned} \int_{\frac{2}{3}}^1g_3(x)f''(x)\mathrm{d}x &=g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1-\int_{\frac{2}{3}}^1g_3'(x)f'(x)\mathrm{d}x\\ &=g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1-g_3'(x)f(x)\bigg|_{\frac{2}{3}}^1+\int_{\frac{2}{3}}^1g_3''(x)f(x)\mathrm{d}x \end{aligned}\tag{3} $$ Let $g(x)$ be the piecewise differentiable function, $g_1(x)$, $g_2(x)$ and $g_3(x)$ is polynomial. $$g(x)=\begin{cases} g_1(x),\quad x\in[0,\tfrac{1}{3})\\ g_2(x),\quad x\in[\tfrac{1}{3},\tfrac{2}{3})\\ g_3(x),\quad x\in[\tfrac{2}{3},1] \end{cases}$$ Since $\textstyle\int_{\frac{1}{3}}^{\frac{2}{3}}f(x)\mathrm{d}x=0$, Let $g_1''(x)=g_3''(x)=c$, $g_2''(x)=k$, So $g_1(x),g_2(x),g_3(x)$ is quadratic polynomial. $$\begin{cases} g_1(x)=cx^2+a_1x+a_0,& x\in[0,\tfrac{1}{3})\\ g_2(x)=kx^2+b_1x+b_0,& x\in[\tfrac{1}{3},\tfrac{2}{3})\\ g_3(x)=cx^2+c_1x+c_0,& x\in[\tfrac{2}{3},1] \end{cases}$$ Let, $$g_1(x)f'(x)\bigg|_0^{\frac{1}{3}}+g_3(x)f'(x)\bigg|_{\frac{2}{3}}^1+g_2(x)f'(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}=0$$ we get, $$\tag{4} \begin{cases} g(0)=g(1)=0\\ g_1(\tfrac{1}{3})-g_2(\tfrac{1}{3})=0\\ g_2(\tfrac{2}{3})-g_3(\tfrac{2}{3})=0 \end{cases}\Rightarrow\quad \begin{cases} a_0=c+c_1+c_0=0\\ \tfrac{1}{9}c+\tfrac{1}{3}a_1+a_0=\tfrac{1}{9}k+\tfrac{1}{3}b_1+b_0\\ \tfrac{4}{9}k+\tfrac{2}{3}b_1+b_0=\tfrac{4}{9}c+\tfrac{2}{3}c_1+c_0 \end{cases} $$ Let, $$g_1'(x)f(x)\bigg|_0^{\frac{1}{3}}+g_3'(x)f(x)\bigg|_{\frac{2}{3}}^1+g_2'(x)f(x)\bigg|_{\frac{1}{3}}^{\frac{2}{3}}=0$$ we get, $$\tag{5} \begin{cases} g'(0)=g'(1)=0\\ g_1'(\tfrac{1}{3})-g_2'(\tfrac{1}{3})=0\\ g_2'(\tfrac{2}{3})-g_3'(\tfrac{2}{3})=0 \end{cases}\Rightarrow\quad \begin{cases} a_1=2c+c_1\\ \tfrac{2}{3}c+a_1=\tfrac{2}{3}k+b_1\\ \tfrac{4}{3}k+b_1=\tfrac{4}{3}c+c_1 \end{cases} $$ by (4),(5), it follows that $$a_0=0,a_1=0,k=-2c,b_1=2c,b_0=-\frac{1}{3}c,c_0=c,c_1=-2c$$ Hence, $$g(x)=\begin{cases} g_1(x)=cx^2,& x\in[0,\tfrac{1}{3})\\ g_2(x)=-2cx^2+2cx-\frac{1}{3}c,& x\in[\tfrac{1}{3},\tfrac{2}{3})\\ g_3(x)=cx^2-2cx+c,& x\in[\tfrac{2}{3},1] \end{cases}$$ by $(1)+(2)+(3)$, we have \begin{align*} \int_0^1g(x)f''(x)\mathrm{d}x &=\int_0^{\frac{1}{3}}g_1(x)f''(x)\mathrm{d}x+\int_{\frac{1}{3}}^{\frac{2}{3}}g_2(x)f''(x)\mathrm{d}x+\int_{\frac{2}{3}}^1g_3(x)f''(x)\mathrm{d}x\\ &=\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x+\underbrace{\int_{\frac{1}{3}}^{\frac{2}{3}}g''_2(x)f(x)\mathrm{d}x}_{=0}+\int_{\frac{2}{3}}^1g''_3(x)f(x)\mathrm{d}x\\ &=2c\int_0^1f(x)\mathrm{d}x \end{align*} By Cauchy-Schwarz inequality, \begin{align*} \left(2c\int_0^1f(x)\mathrm{d}x\right)^2&=\int_0^1g(x)f''(x)\mathrm{d}x\\ &\leqslant \int_0^1(g(x))^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x \end{align*} After calculation, We finally get $$\int_0^1(g(x))^2\mathrm{d}x=\frac{c^2}{1215}+\frac{c^2}{135}+\frac{c^2}{1215}=\frac{11c^2}{1215}$$ Therefore, $$4860\left(\int_0^1f(x)\mathrm{d}x \right)^2\leqslant 11\int_0^1\big(f''(x)\big)^2\mathrm{d}x $$