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Suppose $f : [1,3]\to \mathbb R$ such that $-1\leq f(x)\leq 1, x\in [1,3]$, and $$\int _1^3 f(x)\mathrm{d}x = 0. $$ Determine the largest possible value of $$\int _1^3 \frac{f(x)}{x}\mathrm{d}x. $$

I have found that largest value is $\log3$. But what is the role of that integration which is given to be zero. Is there any other way to solve.

Toby Mak
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vqw7Ad
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  • How could you find the answer if you didn't use the fact that $\int_1^3 f(x) dx = 0$? In addition, can you type your question up in MathJax? – Toby Mak Sep 07 '19 at 12:05
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    Using the upper bound of $f$. The integration is boinded above by $\int\limits_{1}^{3}\frac{1}{x}dx=\log 3$ – vqw7Ad Sep 07 '19 at 12:09
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    @J.Doe But if $f(x)=1$ then $\int_1^3 f(x)\ \mathsf dx = 2\ne0$. – Math1000 Sep 07 '19 at 12:14
  • ok. correct. then how we can do? – vqw7Ad Sep 07 '19 at 12:16
  • @Math1000 You don't have to take $f = 1$. The argument shows that we can't go higher than $\log 3$ as long as $-1\leq f\leq 1$. – AlvinL Sep 07 '19 at 12:27
  • yeah .. I thought like that. but then what can be the use of that integration which is given to be zero. – vqw7Ad Sep 07 '19 at 12:29
  • I am considering starting a bounty if no progress is made on proving that $\ln 4/3$ is the greatest possible value. – Toby Mak Sep 08 '19 at 04:07

1 Answers1

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Define $f(x)$ to be $1$ when $1 ≤ x < 2$ and $-1$ when $2 ≤ x ≤ 3$.

Then the integral is equal to:

$$\int_1^2 \frac{f(x)}{x} \mathrm d x + \int_2^3 \frac{f(x)}{x} \mathrm d x$$ $$= \int_1^2 \frac{1}{x} \mathrm d x + \int_2^3 \frac{-1}{x} \mathrm d x$$ $$= [\ln x]_1^2 + [-\ln x]_2^3$$ $$= (\ln 2 - \ln 1) + (- \ln 3 + \ln 2)$$ $$= \ln 4 - \ln 3 = \ln \left( \frac{4}{3} \right)$$

This is most likely the function that gives the greatest value. For constant functions, adjusting where the function 'breaks', and the number in the constant function does not seem to give an answer greater than this.

The requirement that $\int_1^3 f(x) \mathrm d x = 0$ is to make sure that $f(x)$ cannot be equal to $1$ within the interval. Otherwise you can use the argument that the integral is bounded above by $\int_1^3 \ln x \ \mathrm d x = \ln 3$.

Toby Mak
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    Edited my answer. – Toby Mak Sep 07 '19 at 12:36
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    That's indeed the correct solution. It's obvious that the solution is decreasing. To show that the solution must be plus or minus one everywhere, note that you could improve the objective while keeping the integral constraint by subtracting from the right and adding to the left in a neighborhood where the function is strictly in between the bounds. – Bananach Sep 07 '19 at 13:14
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    @Bananach how to prove rigorously that $\ln(4/3)$ is the maximum? – Szeto Sep 07 '19 at 15:37
  • This is what I thought of posting, but I don't know how to prove that it is indeed the maximum. – Math1000 Sep 07 '19 at 16:08
  • @Szeto Claim A: For any non-decreasing function there is a decreasing function with higher objective Proof A: Switch around non decreasing bits of the function. Claim B: For any function that has some values strictly in between 1 and -1 there is a function with values 1 and -1 only with higher objective. Proof B: see above. Claim C: the best function that is 1 first and then -1 is the function given in this answer. Proof C: Trivial. – Bananach Sep 08 '19 at 09:37
  • @Bananach what is ‘objective’? – Szeto Sep 08 '19 at 09:39
  • @Szeto the integral we're trying to maximize – Bananach Sep 08 '19 at 10:02