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Assume $f:[0,\pi]\rightarrow (0,+\infty)$ is continuous. And $$ \int_0^{\pi/2} f(t)\sin t dt =\int_{\pi/2}^\pi f(t)\sin t dt \tag{1} $$ if $$ \int_0^{\pi/2} f(t)\sin t dt < f(\frac{\pi}{2}) \tag{2} $$ then, how to show $$ \int_0^\pi \sqrt{f(t)} dt < \pi \sqrt{f(\frac{\pi}{2})} ~~~~~~? \tag{3} $$

I guess it when I read some paper. I am not sure it is right. And I don't know how to prove it. Seemly, the Holder, Cauchy, Young inequalities are useless for it.

Response to mathworker21 (2023-8-25): I calculated a function liking $\delta^{-2}1_{[0,\delta]}+1_{[\frac{\pi}{4},\frac{\pi}{2})}+c1_{\{\frac{\pi}{2}\}}$. For convenience, let $$ f(t)=\delta^{-2}1_{[0,\delta]}+1_{[\frac{\pi}{4},\frac{\pi}{2}-\epsilon)}+ k(t-\frac{\pi}{2}+\epsilon) 1_{[\frac{\pi}{2}-\epsilon, \frac{\pi}{2}]} $$ First, I have $$ \int_0^{\pi/2} f(t)\sin tdt = -\delta^{-2}\cos\delta+\delta^{-2}-\sin\epsilon+\frac{\sqrt 2}{2}+k-k\cos\epsilon \\ f(\frac{\pi}{2}) = k\epsilon $$ So, $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$ imply $$ k< \frac{ \delta^{-2}\cos\delta - \delta^{-2} +\sin\epsilon -\frac{\sqrt 2}{2} } { 1-\epsilon-\cos\epsilon } \tag{5} $$ On the other hand, we have $$ \int_0^{\pi/2}\sqrt{f(t)}dt = 1+\frac{\pi}{4}-\epsilon+\frac{2}{3k}(k\epsilon)^{3/2} $$ Therefore, $\int_0^{\pi/2}\sqrt{f(t)}dt=\frac{\pi}{2}\sqrt{f(\frac{\pi}{2})}$ imply $$ \sqrt k = \frac{ 1+\frac{\pi}{4}-\epsilon }{ \frac{\pi}{2}\sqrt\epsilon -\frac{2}{3}\epsilon^{3/2} } \tag{6} $$ However, It is not possible to satisfy both conditions (5) and (6) simultaneously. I use Wolfram Mathematica (a software) to get that when $\epsilon=10^{-10},\delta=10^{-10}$, (5) imply $$ k<1.207106781146902863457773*10^{10} $$ (6) imply $$ k=1.291904506901873765116087*10^{10} $$ Of course, I also test other numerical value. For example, $\epsilon=10^{-6},\delta=10^{-6}$ or $\epsilon=10^{-10},\delta=10^{-6}$ and so on. But (5) and (6) can't be satisfied simultaneously.

Besides, I also calculated $$ f(t)=\delta^{-2}1_{[0,\delta]}+1_{[\frac{\pi}{4},\frac{\pi}{2}-\epsilon)}+ k^2(t-\frac{\pi}{2}+\epsilon)^2 1_{[\frac{\pi}{2}-\epsilon, \frac{\pi}{2}]} $$ It also is not the counterexample of statement 2.

The Mathematica code: enter image description here enter image description here

Response to mathworker21 (2023-8-22): I find that $$ f(t) = \delta^{-2}1_{[0,\delta]}(t)+1_{[\frac{\pi}{4},\frac{\pi}{2}]}(t) $$ is not the counterexample of

Statement $2$: For every continuous $f : [0,\frac{\pi}{2}] \to (0,+\infty)$ satisfying $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$, we have $\int_0^{\pi/2} \sqrt{f(t)}dt < \frac{\pi}{2}\sqrt{f(\frac{\pi}{2})}$.

For this $f(t)$ there always be $\int_0^{\pi/2} \sqrt{f(t)}dt < \frac{\pi}{2}\sqrt{f(\frac{\pi}{2})}$. However, $$ \int_0^{\pi/2} f(t)\sin(t)dt =-\delta^{-2}\cos\delta +\delta^{-2}+\cos\frac{\pi}{4} ~~~~~~~ f(\frac{\pi}{2})=1 $$ for any $\delta\in (0,\frac{\pi}{4})$, there is $$ -\delta^{-2}\cos\delta +\delta^{-2}+\cos\frac{\pi}{4} >1 $$

enter image description here

Response to mathworker21 (2023-8-11): Could you explain why it is equivalent to showing $$ \left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{f(t)}dt\right)^2 \le \int_0^{\pi/2} f(t)\sin(t)dt ~~? \tag{4} $$ In fact, I also don't understand Ryszard Szwarc's comment.

Besides, I calculate your example, when $\delta>0$ is sufficiently small, it is not false. For $$ f(t) = \delta^{-1/2}1_{[0,\delta]}(t)+1_{[\frac{\pi}{4},\frac{\pi}{2}]}(t) $$ I have $$ \int_0^{\pi/2} \sqrt{f(t)}dt= \int_0^\delta \sqrt{\frac{1}{\sqrt\delta}}dt + \int_{\pi/4}^{\pi/2} dt =\delta^{3/4}+\frac{\pi}{4} $$ Therefore, the left part of (4) is $$ L=\frac{4}{\pi^2} \delta^{3/2} +\frac{2}{\pi}\delta^{3/4}+\frac{1}{4} $$ On the other hand, the right part of (4) is $$ R= \int_0^{\pi/2} f(t)\sin t dt = \int_0^\delta \frac{1}{\sqrt\delta} \sin t dt + \int_{\pi/4}^{\pi/2} \sin t dt = \frac{1}{\sqrt\delta}-\frac{1}{\sqrt\delta}\cos\delta+\cos\frac{\pi}{4} $$ And $$ \lim_{\delta\rightarrow 0^+} L =\frac{1}{4} ~~~~~~ \lim_{\delta\rightarrow 0^+} R =\cos\frac{\pi}{4}=\frac{\sqrt 2}{2} $$ So, when $\delta>0$ is sufficiently small, there is $L\le R$.

But, I use Mathematica to get the grapha of $L-R$, there is $L>R$ when $\delta$ near $\pi/4$.

$x$ is the $\delta$ enter image description here

PS(2023-8-8): From two aspects, I think it is right. First, by the rearrangement inequality, I feel it is right. But it is not general rearrangement, the maximum value of $f$ should be placed near $\frac{\pi}{2}$ in the rearrangement. But I still can't give a detailed proof up to now.

On the other hand, I write a program to verify it. I approximate the integral by summation. There is not counter-example. The Python code is as follows:

import math
import random

i=0
while i<10000:
    i=i+1

    #生成一个随机函数  (Generates a random function)
    f_list=[]
    for i0 in range(0,200):
        f_list.append(random.random())

    #计算左边积分  (Compute the left-hand integral of (1) )
    L=0
    for i1 in range(0,100):
        L=L+f_list[i1]*math.sin(i1*math.pi/200)*(math.pi/200)

    #计算右边积分  (Compute the right-hand integral of (1) )
    R=0
    for i2 in range(100,200):
        R=R+f_list[i2]*math.sin(i2*math.pi/200)*(math.pi/200)

    #调整两边积分的大小  (Adjust the size of the integral on both sides of (1) )
    d=L/R
    for i3 in range(100,200):
        f_list[i3]=f_list[i3]*d

    #判断是否小于f(pi/2)  (Determine if it is less than f(pi/2))
    if L>= f_list[100]:
        i=i-1
        continue   #跳过循环剩下部分  (Skip the rest of the loop)

    #计算最后一个式子中左边的积分  (Calculate the integral on the left of the (3) )
    LL=0
    for i4 in range(0,200):
        LL=LL+math.sqrt(f_list[i4])*(math.pi/200)

    #如果是反例,就打印出来  (If it's a counterexample, print it out)
    if LL >= math.pi*math.sqrt(f_list[100]):
        print(LL,math.pi*math.sqrt(f_list[100]),LL-f_list[100])

print("End")
$\endgroup$
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  • $\begingroup$ Where does this come from? Why do you think that the inequality holds? $\endgroup$
    – Martin R
    Aug 7, 2023 at 16:10
  • $\begingroup$ @MartinR I just thought of it, not where I saw it. At beginning, I just feel it maybe right. I don't know how to explain this feeling. Later, I feel it can be proved by some special rearrangement. Besides, I write a program to verify it. I have add the detail in problem. Thanks. $\endgroup$
    – Enhao Lan
    Aug 8, 2023 at 7:57
  • $\begingroup$ It suffices to show that $$\int\limits_0^{\pi/2}\sqrt{f(t)}\,dt<{\pi\over 2}\sqrt{f(\pi/2)}$$ Both inequalities are satisfied if the maximal value is attained at $\pi/2,$ and the function is not constant on the interval $[0,\pi/2].$ So potential counterexample should attain the maximal value somewhere in $(0,\pi/2).$ $\endgroup$ Aug 8, 2023 at 11:00
  • $\begingroup$ @RyszardSzwarc I don't know why it suffices to show $\int\limits_0^{\pi/2}\sqrt{f(t)}\,dt<{\pi\over 2}\sqrt{f(\pi/2)}$. Could you explain it? Thanks. $\endgroup$
    – Enhao Lan
    Aug 11, 2023 at 12:59

1 Answer 1

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Your question is equivalent to showing $$\left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{f(t)}dt\right)^2 \le \int_0^{\pi/2} f(t)\sin(t)dt$$ for any measurable $f : [0,\frac{\pi}{2}] \to (0,+\infty)$.

But this is of course false. For example, one can take $$f(t) = \delta^{-2}1_{[0,\delta]}(t)+1_{[\frac{\pi}{4},\frac{\pi}{2}]}(t)$$ for a sufficiently small $\delta > 0$ (probably $\delta := 10^{-100}$ suffices).

If you want a concrete counterexample to your original problem, define $f(t)$ on $[0,\frac{\pi}{2}]$ as follows and then define $f(t) := f(\pi - t)$ for $t \in [\frac{\pi}{2},\pi]$: for $0 \le t \le \delta$, let $f(t) = \delta^{-2}$, then drop $f$ linearly from $\delta^{-2}$ to $0$ over $\delta \le t \le \delta+\delta^{10}$, then keep $f$ at $0$ for $\delta+\delta^{10} \le t \le \frac{\pi}{4}-\delta^{10}$, then increase $f$ linearly from $0$ to $1$ over $\frac{\pi}{4}-\delta^{10} \le t \le \frac{\pi}{4}$, then keep $f$ at $1$ for $\frac{\pi}{4} \le t \le \frac{\pi}{2}$.

Of course, there could be a cleaner counterexample.


I now show the equivalence.

Statement $1$: For every continuous $f : [0,\pi] \to (0,+\infty)$ satisfying $\int_0^{\pi/2} f(t)\sin(t)dt = \int_{\pi/2}^\pi f(t)\sin(t)dt$ and $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$, we have $\int_0^\pi \sqrt{f(t)}dt < \pi \sqrt{f(\frac{\pi}{2})}$.

Statement $2$: For every continuous $f : [0,\frac{\pi}{2}] \to (0,+\infty)$ satisfying $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$, we have $\int_0^{\pi/2} \sqrt{f(t)}dt < \frac{\pi}{2}\sqrt{f(\frac{\pi}{2})}$.

Claim: Statement $1$ is equivalent to Statement $2$.

Proof:

Let's first show Statement $2$ implies Statement $1$.

We are given a continuous $f : [0,\pi] \to (0,+\infty)$ satisfying $\int_0^{\pi/2} f(t)\sin(t)dt = \int_{\pi/2}^\pi f(t)\sin(t)dt$ and $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$. By the second condition, we have $\int_0^{\pi/2} \sqrt{f(t)}dt < \frac{\pi}{2}\sqrt{f(\frac{\pi}{2})}$. Letting $g(t) = f(\pi-t)$ for $0 \le t \le \frac{\pi}{2}$, we have that $g: [0,\frac{\pi}{2}] \to (0,+\infty)$ is a continuous function satisfying (by the first condition and that $g(\frac{\pi}{2}) = f(\frac{\pi}{2})$) the inequality $\int_0^{\pi/2} g(t)\sin(t)dt < g(\frac{\pi}{2})$. Therefore, by Statement $2$ applied to $g$, we have $\int_0^{\pi/2} \sqrt{g(t)}dt < \frac{\pi}{2}\sqrt{g(\frac{\pi}{2})}$, which is equivalent to $\int_{\pi/2}^\pi \sqrt{f(t)}dt < \frac{\pi}{2}\sqrt{f(\frac{\pi}{2})}$. Adding the two inequalities we've obtained gives $\int_0^\pi \sqrt{f(t)}dt < \pi\sqrt{f(\frac{\pi}{2})}$.

Let's now show Statement $1$ implies Statement $2$.

We are given a continuous $f : [0,\frac{\pi}{2}] \to (0,+\infty)$ satisfying $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$. Extend $f$ to $[0,\pi]$ by definition $f(t) := f(\pi-t)$ for $\frac{\pi}{2} \le t \le \pi$. This extended function $f$ is continuous on $[0,\pi]$, still takes values in $(0,+\infty)$, and satisfies $\int_0^{\pi/2} f(t)\sin(t)dt = \int_{\pi/2}^\pi f(t)\sin(t)dt$ and $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$. Therefore, by Statement $1$, we obtain $\int_0^\pi \sqrt{f(t)}dt < \pi\sqrt{f(\frac{\pi}{2})}$, but $\int_0^\pi \sqrt{f(t)}dt = 2\int_0^{\pi/2} \sqrt{f(t)}dt$ by the definition of the extended $f$. $\square$

Statement $3$: For every measurable $f : [0,\frac{\pi}{2}] \to (0,+\infty)$, we have $$\left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{f(t)}dt\right)^2 \le \int_0^{\pi/2} f(t)\sin(t)dt.$$

Claim: Statement $2$ is equivalent to Statement $3$.

Proof:

Let's first show Statement $3$ implies Statement $2$.

We are given a continuous function $f : [0,\frac{\pi}{2}] \to (0,+\infty)$ satisfying $\int_0^{\pi/2} f(t)\sin(t)dt < f(\frac{\pi}{2})$. By Statement $3$, we know $\int_0^{\pi/2} \sqrt{f(t)}dt \le \frac{\pi}{2}\sqrt{\int_0^{\pi/2} f(t)\sin(t) dt}$. By our assumption, we obtain $\int_0^{\pi/2} \sqrt{f(t)}dt < \frac{\pi}{2}\sqrt{f(\frac{\pi}{2})}$, as desired.

Now let's show Statement $2$ implies Statement $3$. It will be easier to prove the contrapositive, namely that a counterexample to Statement $3$ can produce a counterexample to Statement $2$.

So, we are given a measurable function $f : [0,\frac{\pi}{2}] \to (0,+\infty)$ with $\left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{f(t)}dt\right)^2 > \int_0^{\pi/2} f(t)\sin(t)dt$. Take $\epsilon > 0$ so that $\left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{f(t)}dt\right)^2 > \epsilon+ \int_0^{\pi/2} f(t)\sin(t)dt$. Since continuous functions are dense in measurable functions, we can take some continuous function $g : [0,\frac{\pi}{2}] \to (0,+\infty)$ so that $\int_0^{\pi/2} |f(t)-g(t)|dt \le \frac{1}{10}\epsilon$. Then this function $g$ satisfies $\left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{g(t)}dt\right)^2 > \frac{2}{3}\epsilon+ \int_0^{\pi/2} g(t)\sin(t)dt$.

Finally, by modifying the values of $g$ extremely close to $\frac{\pi}{2}$, let $h : [0,\frac{\pi}{2}] \to (0,+\infty)$ be a continuous function with $h(\frac{\pi}{2}) = \left(\frac{1}{\pi/2} \int_0^{\pi/2} \sqrt{h(t)}dt\right)^2$ and $\int_0^{\pi/2} |g(t)-h(t)|dt < \frac{1}{10}\epsilon$. We claim this $h$ serves as a counterexample to Statement $2$. Clearly we don't have (the strict inequality) $\int_0^{\pi/2} \sqrt{h(t)}dt < \frac{\pi}{2}\sqrt{h(\frac{\pi}{2})}$, so it just suffices to verify $\int_0^{\pi/2} h(t)\sin(t)dt < h(\frac{\pi}{2})$. But this follows from the facts that $\left(\frac{1}{\pi/2}\int_0^{\pi/2} \sqrt{h(t)}dt\right)^2 > \frac{1}{3}\epsilon+ \int_0^{\pi/2} h(t)\sin(t)dt > \int_0^{\pi/2} h(t)\sin(t)dt$ and $h(\frac{\pi}{2}) = \left(\frac{1}{\pi/2} \int_0^{\pi/2} \sqrt{h(t)}dt\right)^2$. $\square$

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  • $\begingroup$ Thanks your answer. But I have some query which is a little long. Therefore, I write it in above problem. Could you help me? Thanks. $\endgroup$
    – Enhao Lan
    Aug 11, 2023 at 12:57
  • $\begingroup$ @lanse7pty It's best to ask a new question. Link this question as context in your new post. $\endgroup$
    – Bumblebee
    Aug 11, 2023 at 13:02
  • $\begingroup$ @lanse7pty I sincerely apologize; it should be $\delta^{-2}$ instead of $\delta^{-1/2}$. I just updated my answer. I'll explain the equivalence later, I need to tend to my mother now. $\endgroup$ Aug 11, 2023 at 18:28
  • $\begingroup$ Thanks very much. You solve a problem that had been bothering me for a long time. Although the results were not what I expected, but math is so. This problem has a equivocal geometric background. Let me think whether I am wrong or some thing I miss. Thanks again. $\endgroup$
    – Enhao Lan
    Aug 12, 2023 at 12:05
  • $\begingroup$ @lanse7pty No problem, nice question. Let me know if you want me to add the equivalence between Statement $2$ and the statement at the beginning of my answer. $\endgroup$ Aug 12, 2023 at 16:06

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