Let $f :[0 \; 1] \rightarrow R $ be a continuous function satisfying
$ \sqrt{xy}(f(x) + f(y)) \leq 1 \; \forall x,y \in [0\; 1]$ ....(1)
Show that
$\int_0^1 f(t) dt \leq \frac{\pi}{2} $
.... (2)
Is there any continuous function satisfying the inequality (1) and achiveing the equality (2)
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Where did this problem arise? $\endgroup$– Yemon ChoiCommented Apr 5, 2023 at 14:45
-
$\begingroup$ Then I don't think MathOverflow is an appropriate place for other people to be helping students with assignments $\endgroup$– Yemon ChoiCommented Apr 5, 2023 at 14:47
-
$\begingroup$ The point of challenges is for you to attempt them, and then to ask the person who posed the challenge when you get stuck. Otherwise, they would not be challenges. The reward is in the attempt, not in being able to find someone else's answer. $\endgroup$– Yemon ChoiCommented Apr 5, 2023 at 14:55
-
$\begingroup$ This site is for research level questions. For general questions in mathematics see math.stackexchange.com $\endgroup$– GH from MOCommented Apr 5, 2023 at 14:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
In the inequality $\sqrt{xy}(f(x)+f(y))\leq 1$ set $y=1-x$. Then $$f(x)+f(1-x)\leq\frac{1}{\sqrt{x(1-x)}}.$$
Integrate from $0$ to $1$ and obtain $$2\int_0^1f(x)dx\leq\int_0^1\frac{dx}{\sqrt{x(1-x)}}=\pi.$$ So $$\int_0^1f(x)dx\leq\frac{\pi}{2}.$$
answered Apr 5, 2023 at 14:49
-
$\begingroup$ Since it is known that this is a challenge problem, and that the user is not interested, I think the question should not be answered here. $\endgroup$– LSpiceCommented Apr 5, 2023 at 18:50
-
$\begingroup$ @LSpice: I do not really care about the circumstances that you mention. Do you know what is the best possible estimate, and an extremal function? $\endgroup$ Commented Apr 6, 2023 at 13:32
-
$\begingroup$ @Yemon Choi: Thanks for the correction. $\endgroup$ Commented Apr 6, 2023 at 18:54