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Find the minimum of the value $k$, such have $$\int_{0}^{1}f^2(x)dx\le k\left(\int_{0}^{1}f(x)dx\right)^2$$ for any integrable function $f(x)$,and $1\le f(x)\le 2,x\in(0,1)$

My idea: use Cauchy-Schwarz inequality we have $$\int_{0}^{1}f^2(x)dx\ge\left(\int_{0}^{1}f(x)dx\right)^2$$

I know this can't usefull to solve my problem. Thank you

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  • $\begingroup$ Your finite dimensional analog is to find $\sup_{x\in \mathbb{R}^n}\frac{||x||^2}{<x,1>^2}$ $\endgroup$ – Samrat Mukhopadhyay May 8 '14 at 15:11
  • $\begingroup$ @becko,division? at last I can't find it,But Thank you $\endgroup$ – math110 May 8 '14 at 15:13
  • $\begingroup$ @Samrat That supremum is infinite, because by choosing an $x$ orthogonal or nearly orthogonal to $1$, you get that quantity to be as large as you want. You want to restrict the supremum to $x\in[1,2]^n$. $\endgroup$ – Mario Carneiro May 8 '14 at 15:26
  • $\begingroup$ @MarioCarneiro correct, correct. $\endgroup$ – Samrat Mukhopadhyay May 8 '14 at 15:46
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1). Since $(2-f(x))(f(x)-1)\geq0$, so

$$\int_0^1\Big(2-f(x)\Big)\Big(f(x)-1\Big)\,dx\geq0$$

that is

$$3\int_0^1f(x)\,dx-2\geq \int_{0}^{1}f^2(x)\,dx$$

2). Set $t=\int_0^1f(x)\, dx$, $1\leq t\leq2$. We want

$$kt^2\geq 3t-2\qquad1\leq t\leq2$$

that is $kt+\dfrac2t\geq3$. By A-G inequality, we get $$k\geq\dfrac98$$ On the other hand $$\dfrac98t^2- \big(3t-2\big)=\dfrac98\Big(t-\frac43\Big)^2\geq0$$ so $k=\dfrac98$ is sufficient!

3) If $\int_{0}^{1}f^2(x)dx=\dfrac98\left(\int_{0}^{1}f(x)dx\right)^2$, we have

$$t=\int_{0}^{1}f(x)dx=\frac43\qquad \int_{0}^{1}f^2(x)\,dx=2$$

en

\begin{equation}f(x)=\begin{cases}1& 0\leq x\leq\frac23\\2&\frac23\lt x\leq1\end{cases}\end{equation}

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Let $k=\sup_{f:[0,1]\to[1,2]}\frac{\int f^2}{(\int f)^2}.$ The function $f(x)=2$ if $x\le\frac13$ and $f(x)=1$ otherwise provides the lower bound $k\ge\frac98$. I claim that $k=\frac98$. Let $\epsilon(x)=\epsilon\,\delta_a(x_0)$ be a rectangular bump function at $x_0$ with width $a\ll1$ and integral $\epsilon\ll a$.

$$\Big(\int f(x)+\epsilon(x)\,dx\Big)^2=\Big(\int f\Big)^2+2\epsilon\int f+\epsilon^2\\ \int (f(x)+\epsilon(x))^2\,dx=\int f^2+2\epsilon f(x_0)+\epsilon^2/a$$

Thanks to our choices $\epsilon\ll a\ll1$, we can ignore the final term of each expansion, so we are left with the (unsurprising) conclusion that given any $f$, we can increase the quantity $\int f^2/(\int f)^2$ by moving a chunk of $f$ away from its average value. Thus any extremal $f$ must have every point as far from its average value as possible, that is, we must have $f:[0,1]\to\{1,2\}$. The integrals of such functions are driven entirely by the measure of $f^{-1}(\{2\})$; labeling this quantity as $z$, we have

$$\int f^2=4z+(1-z)=3z+1\\ \Big(\int f\Big)^2=(2z+(1-z))^2=(z+1)^2,$$

and the quantity $\frac{3z+1}{(z+1)^2}$ attains a unique maximum in $[0,1]$ at $z=\frac13$, with value $\frac98$ (and leading to our original example for a lower bound). Thus $k=\frac98$.

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  • $\begingroup$ It's nice! Thank you +1 $\endgroup$ – math110 May 9 '14 at 7:54

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