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

[Ackbach]

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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[Ackbach]

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):

Spoiler

(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.