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let $f$ on $[a,b]$ two continuously differentiable functions,such

$$f(a)=f(b)=0, f'(a)=1,f'(b)=0,b>a>0$$ show that $$\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}$$

My idea: use Cauchy-Schwarz inequality $$\int_{a}^{b}[f''(x)]^2dx\cdot\int_{a}^{b}g^2(x)dx\ge\left(\int_{a}^{b}f''(x)g(x)dx\right)^2$$ How find the $g(x)$?

then I think

\begin{align} \int_{a}^{b}f''(x)g(x)dx &=\int_{a}^{b}g(x)df'(x)=f'(x)g(x)|_{a}^{b}-\int_{a}^{b}f'(x)g(x)dx\\ &=f'(b)g(b)-f'(a)g(a)-\int_{a}^{b}g(x)df(x)\\ &=-g(a)+\int_{a}^{b}g'(x)dF(x)\\ &=-g(a)+F(x)g'(x)|_{a}^{b}-\int_{a}^{b}F(x)g''(x)\\ &=-g(a)+F(b)g'(b)-F(a)g'(a)-\int_{a}^{b}F(x)g''(x)dx \end{align} where $F(x)=\int_{a}^{x}f(t)dt$

Now following I can't find the $g(x)$? can you help me or use other methods solve it? Thank you

and some hours ago: I ask this integral inequality:How prove this $\int_{a}^{b}f^2(x)dx\le (b-a)^2\int_{a}^{b}[f'(x)]^2dx$

Community
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math110
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    I believe you should minimize the functional: $J(g)=\frac{(\int_{a}^{b} f''(x)g(x) dx)^2}{\int_{a}^{b} g^2(x) dx}$ with methods from calculus of variations. I am a bit rusty here, but you shoudl take derivative of $J$ wrt $g$, and equate it to zero. – MathematicalPhysicist Sep 20 '13 at 08:30

4 Answers4

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For the function $$f(x) = \frac{(x-a)(b-x)^2}{(b-a)^2}$$ that integral is exactly $$\frac{4}{b-a}.$$ If $g$ is twice continuously differentiable and $g(a)=g(b)=g'(a)=g'(b)=0$ then

$$\begin{eqnarray} \int_a^b(f''+g'')^2 &\geq& \frac{4}{b-a} +2\int_a^b f'' g''\\ &=& \frac{4}{b-a} - 2\int_a^b f^{(3)}g' \\ &=& \frac{4}{b-a} + 2\int_a^b f^{(4)}g \\ &=& \frac{4}{b-a} \end{eqnarray} $$

This proves the general case since any such function $h$ can be written as $h = f + (h - f)$.

WimC
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  • Is there any intuition behind this particular $f$? – Tomás Sep 29 '13 at 19:34
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    @Tomás Yes there is. First suppose that the extremal function is smooth, then use calculus of variations to derive its properties. (Perturb with functions such as $g$ in the answer). This will show that $f^{(4)}=0$ so it must be a polynomial of degree at most three. – WimC Sep 29 '13 at 20:02
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I found $$g(x)=6x-2a-4b$$ because $$\int_{a}^{b}[f''(x)]^2dx\int_{a}^{b}(6x-2a-4b)^2dx\geqslant\left(\int_{a}^{b}(6x-2a-4b)f''(x)dx\right)^2$$

and $$\int_{a}^{b}(6x-2a-4b)^2dx=4(b-a)^3$$ $$\left(\int_{a}^{b}(6x-2a-4b)f''(x)dx\right)^2=[4(b-a)]^2$$

Shine Mic
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Let $g(x)$ be a polynomial, by integration by parts, \begin{align*} \int_a^bg(x)f''(x)\mathrm{d}x &=g(x)f'(x)\bigg|_a^b-\int_a^bg'(x)f'(x)\mathrm{d}x\\ &=-g(a)+\underbrace{\int_a^bg''(x)f(x)\mathrm{d}x}_{=0} \end{align*} Only let $g''(x)=0\Rightarrow g(x)=c_1x+c_0$. by the Cauchy-Schwarz inequality, $$\int_a^b\big(f''(x)\big)^2\mathrm{d}x\int_a^bg^2(x)\mathrm{d}x\geqslant\left(\int_a^bg(x)f''(x)\mathrm{d}x\right)^2\longrightarrow\text{Constants}$$ We have, $$\int_a^b\big(f''(x)\big)^2\mathrm{d}x\geqslant \frac{g^2(a)}{\int_a^bg^2(x)\mathrm{d}x}=\frac{4}{b-a}$$ Substituting, $$3c_1(b-a)(c_1a+c_0)^2=4\big[(c_1b+c_0)^3-(c_1a+c_0)^3\big]$$ Simplification, we get $$\big((a+2b)c_1+3c_0\big)^2=0\Rightarrow (a+2b)c_1=-3c_0$$ So, we get, $$g(x)=c_1x-\frac{1}{3}(a+2b)c_1$$ Inside,$c_1$ is a constant. $\Box$

lingdanda
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  • Could you help me to upvote my bonutied questions, which have been flop for so long ?? –  Feb 28 '21 at 03:52
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This is the special case of Schoenberg theorem on cubic splines (piecewise polynomials), saying that the cubic spline minimizes the curvature, see these notes, Theorem on p.5.

Since there are no conditions for your function inside the interval, the extremal spline transforms into cubic polynomial. So what you really need is to find cubic polynomial that satisfies boundary conditions.

The proof of Schoenberg theorem is pretty much the same as you outlined in your post.

leshik
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