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Let $f,g$ be two continuous positive functions over $[a,b]$

Let $m_1$ and $M_1$ be the minimum and maximum of $f$

Let $m_2$ and $M_2$ be the minimum and maximum of $g$

Prove that $$\sqrt{\int_a^bf^2 \int_a^b g^2}\leq \frac{1}{2}\left(\sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}\right)\int_a^bfg$$

I can't make any significant progress on this one...

Thanks for any hint.

Community
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Gabriel Romon
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    This looks like a variation of the Polya-Szego inequality, but I don't know how to prove it. Maybe you can search on that and find something? – MPW Jul 27 '14 at 19:00
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    Use $f^2(x) = f(x) \cdot g(x) \cdot \frac{f(x)}{g(x)} \le f(x) \cdot g(x) \cdot \frac{M_1}{m_2}$ and a similar inequality for $g^2(x)$. – Hans Engler Jul 27 '14 at 19:05
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    @MPW You are right, http://rgmia.org/papers/v7n1/ReverseSchwarz.pdf. – Krokop Jul 27 '14 at 19:05
  • @Krokop : Thank you for the link ! Great document ! – pitchounet Jul 27 '14 at 19:32
  • @HansEngler I think it's not enough... – Gabriel Romon Jul 28 '14 at 08:04
  • Good question! Does anyone know if there's a similar inequality for functions which are Lebesgue measurable w.r.t. a measure $\mu$ and a.s. bounded away from $0, \infty?$ (so $\mu$-a.s. positive and bounded functions?) – Mathguest May 02 '20 at 12:49

3 Answers3

17

We use continuity of $f$ and $g$, to prove that the integral inequality: $$\displaystyle \int_a^b f^2\int_a^bg^2 \le \dfrac{1}{4}\left(\sqrt{\dfrac{M_1M_2}{m_1m_2}}+\sqrt{\dfrac{m_1m_2}{M_1M_2}}\right)^2\left(\int_a^bfg\right)^2$$

follows from proving the discrete case of the inequality, taking $a_k = f(a+\frac{b-a}{n}k)$ and $b_k = g(a+\frac{b-a}{n}k)$, for $k=1,\cdots,n$.

$$\sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2 \le \left(\dfrac{M+m}{2\sqrt{Mm}}\sum_{k=1}^n a_kb_k\right)^2$$

where, $\displaystyle M = {\dfrac{M_1}{m_2}} \ge \dfrac{a_k}{b_k} \ge {\dfrac{m_1}{M_2}} = m$, for all $1 \le k \le n$.

Using, $\displaystyle \left(\dfrac{a_k}{b_k} - M\right)\left(\dfrac{a_k}{b_k} - m\right) \le 0 \implies a_k^2+Mmb_k^2 \le (M+m)a_kb_k$

Thus, $$\sum_{k=1}^na_k^2+Mm\sum_{k=1}^nb_k^2 \le (M+m)\sum_{k=1}^na_kb_k$$

Using Am-Gm inequality,

$$2\left(\sum_{k=1}^na_k^2\right)^{1/2}\left(Mm\sum_{k=1}^nb_k^2\right)^{1/2} \le \sum_{k=1}^na_k^2+Mm\sum_{k=1}^nb_k^2 \le (M+m)\sum_{k=1}^na_kb_k$$

Dividing both sides by $n$ and taking limit as $n \to \infty$,

$$\lim\limits_{n\to \infty}\left(\dfrac{1}{n}\sum_{k=1}^na_k^2\right)^{1/2}\left(\dfrac{1}{n}\sum_{k=1}^nb_k^2\right)^{1/2} \le \lim\limits_{n\to \infty}\dfrac{M+m}{2\sqrt{Mm}}\dfrac{1}{n}\sum_{k=1}^na_kb_k$$

$$\implies \displaystyle \sqrt{\int_a^b f^2\int_a^bg^2} \le \dfrac{1}{2}\left(\sqrt{\dfrac{M_1M_2}{m_1m_2}}+\sqrt{\dfrac{m_1m_2}{M_1M_2}}\right)\int_a^bfg$$

user547265
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r9m
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    I kept this window open too long and didn't see your answer appear. Good answer. (+1) I think we both used the same idea; that is, I think $(4)$ in my answer is similar to your inequality with $M$ and $m$. – robjohn Jul 28 '14 at 21:30
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    Integrating both sides of $\displaystyle f^2 + Mm g^2 \le (M+m)fg \implies \int_0^1 f^2 + Mm \int_0^1 g^2 \le (M+m)\int_0^1 fg$ and applying Am-Gm to the LHS, establishes the required inequality without having to use the continuity (used to move from the discrete to the continuous case) – r9m Jul 28 '14 at 22:42
  • and avoids the scaling that I did. – robjohn Jul 28 '14 at 23:00
  • Why do we require $\displaystyle M = {\dfrac{M_1}{m_2}} \ge \dfrac{a_k}{b_k} \ge {\dfrac{m_1}{M_2}} = m$, for $1 \le k \le n$? – user547265 Jun 11 '18 at 22:34
  • @user547265 it's not a requirement, it follows from considering the two inequalities $M_1 \ge a_k \ge m_1$ and $M_2 \ge b_k \ge m_2$. – r9m Jun 11 '18 at 22:43
  • Hmm, I don't know why do we consider $M_1 \ge a_k \ge m_1$ and $M_2 \ge b_k \ge m_2$. I understand the rest of the proof, but this $\displaystyle M = {\dfrac{M_1}{m_2}} \ge \dfrac{a_k}{b_k} \ge {\dfrac{m_1}{M_2}} = m$ looks like a magic. I want to understand that. Would you mind elaborating? – user547265 Jun 11 '18 at 22:54
  • The key step is the rearrangement inequality used, $\displaystyle (M+m)ab \ge a^2 + Mmb^2 \quad \cdots (1)$. Suppose we were trying to prove the inequality for $n=1$ case (only one term in the inner product), we would try to find constants $t,s > 0$ s.t., $ab \ge ta^2 + sb^2$ holds. How would you propose we choose these $s,t$? – r9m Jun 11 '18 at 23:30
  • @r9m Nice answer, upvoted! I see that you used continuity and Riemann integrability in your proof while passing to the limit. Do you know if there's a similar inequality for functions which are Lebesgue measurable w.r.t. a measure $\mu$ and a.s. bounded away from $0, \infty?$ (so $\mu$-a.s. positive and bounded functions?) Maybe then we can first prove it for step functions, and then pass to all $L^1$ -functions? – Mathguest May 02 '20 at 12:51
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    @LearningMath robjohn's proof below covers the case for $L^p$ functions. You see in both rob's proof and mine we are using eqn $(4)$ which is a pointwise inequality. After that you can integrate both sides w.r.t. any measure (as long as the integrals make sense) and get the inequalities. – r9m May 02 '20 at 13:06
  • @r9m thanks a lot, yes I didn't see it before, will check now, thanks! – Mathguest May 02 '20 at 13:35
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Here is a proof of a reverse Hölder inequality proven in a manner very similar to the proof of the reverse Cauchy-Schwarz inequality in my other answer. In what follows, $p,q\gt1$ and $\frac1p+\frac1q=1$. The result requested in the question is simply the case $p=q=2$.

By scaling, which does not change the ratio of maximum to minimum, we can assume $$ \int_a^bf(x)^p\,\mathrm{d}x=\int_a^bg(x)^q\,\mathrm{d}x=1\tag{1} $$ Trivially, we have $$ \int_a^bf(x)g(x)\frac{f(x)^{p-1}}{g(x)}\,\mathrm{d}x=\int_a^bf(x)^p\,\mathrm{d}x=1\tag{2} $$ and $$ \int_a^bf(x)g(x)\frac{g(x)^{q-1}}{f(x)}\,\mathrm{d}x=\int_a^bg(x)^q\,\mathrm{d}x=1\tag{3} $$ Since each term is non-negative, $$ \left(\frac{M_f^{p-1}}{m_g}-\frac{f(x)^{p-1}}{g(x)}\right)\left(\frac{M_g^{q-1}}{m_f}-\frac{g(x)^{q-1}}{f(x)}\right)\ge0\tag{4} $$ where $M_f$ and $M_g$ are the suprema and $m_f$ and $m_g$ are the infima of $f$ and $g$.

Integrating $(4)$ against $f(x)g(x)$ using $(2)$ and $(3)$ gives $$ \begin{align} &\left(\frac{M_f^{p-1}}{m_g}\frac{M_g^{q-1}}{m_f}+M_f^{p-2}M_g^{q-2}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\\ &\ge\frac{M_f^{p-1}}{m_g}\frac{M_g^{q-1}}{m_f}\int_a^bf(x)g(x)\,\mathrm{d}x +\int_a^bf(x)^{p-1}g(x)^{q-1}\,\mathrm{d}x\tag{5}\\ &\ge\frac{M_f^{p-1}}{m_g}+\frac{M_g^{q-1}}{m_f}\tag{6}\\ &\ge\left(q\frac{M_f^{p-1}}{m_g}\right)^{\frac1q}\left(p\frac{M_g^{q-1}}{m_f}\right)^{\frac1p}\tag{7} \end{align} $$ Explanation:
$(5)$: $M_f^{p-2}f(x)\ge f(x)^{p-1}$ and $M_g^{q-2}g(x)\ge g(x)^{q-1}$
$(6)$: Integrate $(4)$ against $f(x)g(x)$ and use $(2)$ and $(3)$
$(7)$: Young's Inequality

Therefore, $$ \frac1{p^{\frac1p}q^{\frac1q}}\left[\left(\frac{M_f}{m_f}\right)^{\frac1q}\left(\frac{M_g}{m_g}\right)^{\frac1p}+\left(\frac{m_f}{M_f}\right)^{\frac1p}\left(\frac{m_g}{M_g}\right)^{\frac1q}\right]\int_a^bf(x)g(x)\,\mathrm{d}x\ge1\tag{8} $$ Undoing the scaling used to get $(1)$, we get $$ \begin{align} &\frac1{p^{\frac1p}q^{\frac1q}}\left(\frac{M_f}{m_f}\right)^{\frac{p}{p+q}}\left(\frac{M_g}{m_g}\right)^{\frac{q}{p+q}}\left(1+\frac{m_f}{M_f}\frac{m_g}{M_g}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\\ &\ge\left(\int_a^bf(x)^p\,\mathrm{d}x\right)^{\frac1p}\left(\int_a^bg(x)^q\,\mathrm{d}x\right)^{\frac1q}\tag{9} \end{align} $$

robjohn
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11

By scaling, which does not change the ratio of maximum to minimum, we can assume $$ \int_a^bf(x)^2\,\mathrm{d}x=\int_a^bg(x)^2\,\mathrm{d}x=1\tag{1} $$ Trivially, we have $$ \int_a^bf(x)g(x)\frac{f(x)}{g(x)}\,\mathrm{d}x=\int_a^bf(x)^2\,\mathrm{d}x=1\tag{2} $$ and $$ \int_a^bf(x)g(x)\frac{g(x)}{f(x)}\,\mathrm{d}x=\int_a^bg(x)^2\,\mathrm{d}x=1\tag{3} $$ Since each term is non-negative, $$ \left(\frac{M_f}{m_g}-\frac{f(x)}{g(x)}\right)\left(\frac{M_g}{m_f}-\frac{g(x)}{f(x)}\right)\ge0\tag{4} $$ where $M_f$ and $m_f$ are the maxima and minima of $f$ and $M_g$ and $m_g$ are the maxima and minima of $g$.

Integrating $(4)$ against $f(x)g(x)$ using $(2)$ and $(3)$ gives $$ \begin{align} \left(\frac{M_f}{m_g}\frac{M_g}{m_f}+1\right)\int_a^bf(x)g(x)\,\mathrm{d}x &\ge\frac{M_f}{m_g}+\frac{M_g}{m_f}\\ &\ge2\sqrt{\frac{M_f}{m_g}\frac{M_g}{m_f}}\tag{5} \end{align} $$ Therefore, $$ \frac12\left(\sqrt{\frac{M_f}{m_g}\frac{M_g}{m_f}}+\sqrt{\frac{m_g}{M_f}\frac{m_f}{M_g}}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\ge1\tag{6} $$ Undoing the scaling used to get $(1)$, we get $$ \hspace{-1cm}\frac12\left(\sqrt{\frac{M_f}{m_g}\frac{M_g}{m_f}}+\sqrt{\frac{m_g}{M_f}\frac{m_f}{M_g}}\right)\int_a^bf(x)g(x)\,\mathrm{d}x\ge\sqrt{\int_a^bf(x)^2\,\mathrm{d}x\int_a^bg(x)^2\,\mathrm{d}x}\tag{7} $$

robjohn
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  • (+1) for proving that inequality still holds without assuming continuity ! – r9m Jul 28 '14 at 22:38
  • Looking more closely, $(4)$ does not appear to be as identical to your inequality with $M$ and $m$ as I first thought. They play similar roles, but the proofs are a bit different to account for the distinction. – robjohn Jul 28 '14 at 22:53