MHB How can the integral of a continuous function on [0,1] be minimized?

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The discussion centers on a mathematical problem involving the integral of a continuous function on the interval [0,1]. It presents the inequality that states the double integral of the sum of two function evaluations is greater than or equal to the integral of the absolute value of the function itself. Participants are encouraged to refer to the Problem of the Week guidelines for proper submission. Kiwi receives an honorable mention for an attempt at solving the problem, although it was incorrect. The solution is credited to Kiran Kedlaya and his associates, highlighting the problem's origin from the 2003 Putnam collection.
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Here is this week's POTW:

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Let $f(x)$ be a continuous real-valued function defined on the interval $[0,1]$. Show that
$$\int_0^1 \int_0^1 |f(x)+f(y)| \, dx \, dy \ge \int_0^1 |f(x)| \, dx.$$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 206 - March 8, 2016

Honorable mention goes to Kiwi for a valiant, but not quite correct, attempt. The solution, attributed to Kiran Kedlaya and his associates, follows: (problem taken from 2003 Putnam collection):

(composite of solutions by Feng Xie and David Pritchard) Let $\mu$ denote Lebesgue measure on $[0,1]$. Define
\begin{align*}
E_+ &= \{x \in [0,1]: f(x) \geq 0\} \\
E_- &= \{x \in [0,1]: f(x) < 0\};
\end{align*}
then $E_+$, $E_-$ are measurable and $\mu(E_+) + \mu(E_-) = 1$. Write $\mu_+$ and $\mu_-$ for $\mu(E_+)$ and $\mu(E_-)$. Also define
\begin{align*}
I_+ &= \int_{E_+} |f(x)|\,dx \\
I_- &= \int_{E_-} |f(x)|\,dx,
\end{align*}
so that $\int_0^1 |f(x)|\,dx = I_+ + I_-$.

From the triangle inequality $|a+b| \geq \pm(|a| - |b|)$, we have the inequality
\begin{align*}
&\iint_{E_+ \times E_-} |f(x) + f(y)|\,dx\,dy \\
&\geq
\pm \iint_{E_+ \times E_-} (|f(x)| - |f(y)|)\,dx\,dy \\
&= \pm ( \mu_- I_+ - \mu_+ I_-),
\end{align*}
and likewise with $+$ and $-$ switched. Adding these inequalities together and allowing all possible choices of the signs, we get
\begin{align*}
&\iint_{(E_+ \times E_-) \cup (E_- \times E_+)} |f(x) + f(y)|\,dx\,dy \\
&\geq
\max\left\{ 0, 2 (\mu_- I_+ - \mu_+ I_-), 2 (\mu_+ I_- - \mu_- I_+) \right\}.
\end{align*}
To this inequality, we add the equalities
\begin{align*}
\iint_{E_+ \times E_+} |f(x) + f(y)|\,dx\,dy &= 2 \mu_+ I_+ \\
\iint_{E_- \times E_-} |f(x) + f(y)|\,dx\,dy &= 2 \mu_- I_- \\
-\int_0^1 |f(x)|\,dx &= -(\mu_+ + \mu_-)(I_+ + I_-)
\end{align*}
to obtain
\begin{multline*}
\int_0^1 \int_0^1 |f(x)+f(y)|\,dx\,dy - \int_0^1 |f(x)|\,dx \\
\geq \max\{ (\mu_+ - \mu_-)(I_+ + I_-)+ 2\mu_-(I_- - I_+), \\
(\mu_+ - \mu_-)(I_+ - I_-), \\
(\mu_- - \mu_+)(I_+ + I_-)+ 2\mu_+(I_+ - I_-) \}.
\end{multline*}
Now simply note that for each of the possible comparisons between $\mu_+$ and $\mu_-$, and between $I_+$ and $I_-$, one of the three
terms above is manifestly nonnegative. This yields the desired result.
 

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