# $\int_0^{\pi/2} f(t)\sin t dt =\int_{\pi/2}^\pi f(t)\sin t dt < f(\frac{\pi}{2}) \Rightarrow \int_0^\pi \sqrt{f(t)} dt < \pi \sqrt{f(\frac{\pi}{2})}$

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:

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 $$-\delta^{-2}\cos\delta +\delta^{-2}+\cos\frac{\pi}{4} >1$$

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 $$\delta$$

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")

• Where does this come from? Why do you think that the inequality holds? Aug 7, 2023 at 16:10
• @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. Aug 8, 2023 at 7:57
• 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 $(0,\pi/2).$ Aug 8, 2023 at 11:00
• @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. Aug 11, 2023 at 12:59

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 $$h(\frac{\pi}{2}) = \left(\frac{1}{\pi/2} \int_0^{\pi/2} \sqrt{h(t)}dt\right)^2$$. $$\square$$

• 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. Aug 11, 2023 at 12:57
• @lanse7pty It's best to ask a new question. Link this question as context in your new post. Aug 11, 2023 at 13:02
• @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. Aug 11, 2023 at 18:28
• 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. Aug 12, 2023 at 12:05
• @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. Aug 12, 2023 at 16:06