On the $\max\int_0^1 fg$ where $f$ is fixed

Question

Let $f$ be a continuous function. What is the maximum of $\int_0^1 fg$ among all continuous functions $g$ with $\int_0^1 |g| = 1$?

score 5 · Answer 1 · answered Dec 15 '13 at 15:15

5

Hint: Try concentrating the weight of $g$ around the maximum of $|f|$. What happens then?

answered Dec 15 '13 at 15:15

Pablo Rotondo

2,185

+1. My friend why did you delete your answer here. :-) – Mikasa Dec 27 '13 at 15:59
@B.S. This is going to sound silly, but I find it a bit embarrasing to have the title "Cannabis Equation" as my highest-rated answer. – Pablo Rotondo Dec 29 '13 at 15:15
Please make it back. I voted it to be undeleted. Honestly, your post made me to do that animation :-) – Mikasa Dec 29 '13 at 15:39

ncmathsadist · Accepted Answer · 2013-12-15T15:27:26.940

1

Put $M = \|f\|_\infty$. Note that $$\int_0^1 f(x) g(x)\, dx \le \|f\|_\infty \|g\|_1 = M.$$ Let $\epsilon > 0$. Then choose a point $x$ so $|f(x)| = M$. Wlog, we may assume $f(x) = M$. Choose an interval $I$ so that $f\ge M - \epsilon$ on $I$.

Define $g$ to be $1/|I|$ on $I$ and $0$ off $[f > 0]$; then extend this function continuously onto all of $[0,1]$ so it has values in $[0, 1/|I|]$. Then $$\int_0^1 f(x) g(x)\, dx \ge {1\over |I|}\int_I f(x)g(x)\, dx \ge M - \epsilon.$$ The supremum must be $M$.

If you take $f(x) = x(1-x)$, $x\in [0,1]$, you will see this supremum is not attained.

edited Dec 15 '13 at 15:27

answered Dec 15 '13 at 15:16

ncmathsadist

50,127

1

Is $g$ continuous ? – Gabriel Romon Dec 15 '13 at 15:20
YOu can "round it off" very easily. Just taper the indicator function to zero inside of $[f > 0]$; this this is a neighborhood of $I$, the continuity is not a real issue. – ncmathsadist Dec 15 '13 at 15:21

On the $\max\int_0^1 fg$ where $f$ is fixed

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