Is $\iiint_{[0, 1]^3} \lvert f(x)+f(y)+f(z)\rvert\, dx\, dy\, dz \ge \int_0^1 \lvert f(x)\rvert\, dx$?

Question

$\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Abs[1]{\left\lvert#1\right\rvert}$Question: Let $f(x)$ be a continuous real-valued function defined on the interval $[0, 1]$. Does the following inequality hold? $$\int_{0}^{1}\int_{0}^{1}\dotsi\int_0^1\int_0^1\abs{f(x_{1})+f(x_{2})+\dotsb+f(x_{n})}dx_1 \; dx_{2}\dotsm\;dx_{n} \ge \int_0^1 \abs{f(x)}dx.$$

Even the case $n=3$ would be interesting. This is inspired by the case $n=2$, which was a Putnam problem from 2003, as follows.

Theorem. Let $f(x)$ be a continuous real-valued function defined on the interval $[0, 1]$. Then $$\int_0^1\int_0^1\abs{f(x)+f(y)}dx \; dy \ge \int_0^1 \abs{f(x)}dx.$$

Proof by Kent Merryfield:

Let $P$ be the subset of $[0, 1]$ on which $f\ge 0$ and $N$ the set on which $f < 0$. As is conventional, define $f^+(x) = \max(f(x), 0)$ and $f^-(x) = \max(-f(x), 0)$. Thus, $f = f^+ - f^-$, $|f| = f^+ + f^-$, and $f^+$ equals $0$ everywhere on $N$ while $f^-$ equals zero everywhere on $P$.

Then \begin{align*}\int_0^1 \int_0^1 \abs{f(x) + f(y)}\,dx\,dy &= \int_P \int_P \abs{f(x) + f(y)}\,dx,dy + \int_P \int_N \abs{f(x) + f(y)} \,dx\, dy \\ &+ \int_N \int_P \abs{f(x) + f(y)}\, dx \,dy+ \int_N \int_N \abs{f(x) + f(y)}\, dx\, dy.\end{align*} We tackle these terms one at a time. \begin{align*}\int_P \int_P \abs{f(x) + f(y)}\, dx\, dy &= \int_P \int_P (f(x) + f(y))\,dx \,dy\\ &= \abs P \int_P f(x)\,dx + \abs P \int_P f(y) \,dy = 2\abs P\int_P f^+(x) \,dx\end{align*} where we use the notation $\abs P$ to mean the measure (total net length) of the set $P$.

Similarly, $\int_N \int_N \abs{f(x) + f(y)} \,dx \,dy = 2\abs N\int_N f^-(x) \,dx$.

The other two terms are equal to each other (as shown by interchanging $x$ and $y$). \begin{align*}\int_P \int_N \abs{f(x) + f(y)} \,dx \,dy &= \int_P \int_N \abs{f^+(x) - f^-(y)} \,dx \,dy\\ &\ge \Abs{\int_P \int_N f^+(x) - f^-(y) \,dx \,dy}\\ &= \Abs{ \abs N\int_P f^+(x) \,dx - \abs P\int_N f^-(y) \,dy}\end{align*} If we let $A = \int_P f^+(x) \,dx$, $B = \int_N f^-(x) \,dx$, and $I = \int_0^1 \int_0^1 \abs{f(x) + f(y)} \,dx \,dy$, then we have found that $I \ge 2\abs P A + 2\abs N B + 2\abs{(\abs N A - \abs P B)}$.

For convenience, we now square this: \begin{align*}I^2 &\ge 4\left[(\abs P A + \abs N B)^2 + (\abs N A - \abs P B)^2 + (\text{other positive terms})\right]\\ &\ge 4(\abs P^2A^2 + \abs N^2B^2 + \abs N^2A^2 + \abs P^2B^2)\\ &= 4(\abs P^2 + \abs N^2)(A^2 + B^2).\end{align*} But for real $a$ and $b$, $(a + b)^2 \le 2(a^2 + b^2)$ since $2(a^2 + b^2) - (a + b)^2 = (a - b)^2$.

Hence, $2(\abs P^2 + \abs N^2) \ge (\abs P + \abs N)^2 = 1^2$, since $\abs P + \abs N$ is the measure of the interval $[0, 1]$. Also, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 \abs{f(x)} \,dx\right)^2$. $\square$

Answer 1

Here is the proof using the alternative route.

Let $X$, $Y$ be two independent real-valued random variables such that $EX,EY\ge 0$ and $\min(E|X|,E|Y|)=I$. We want to prove that $E|X+Y|\ge I$. Again, as in both the OP and Iosif's post, we can consider only the case when $X$ is $A$ with probability $P$ and $-B$ with probability $Q$, while $Y$ is $a$ with probability $p$ and $-b$ with probability $q$, where $A,B,a,b\ge 0$.

Then we need to show that the inequality $$ (A+a)Pp+|a-B|pQ+|b-A|Pq+(B+b)Qq<I $$ is impossible. So suppose it holds.

If we estimate each absolute value by what is inside it, we get $$ (A+a)Pp+(a-B)pQ+(b-A)Pq+(B+b)Qq \\ =(ap+bq)+(AP-BQ)(p-q) \\=E|Y|+(EX)(p-q)<I\,, $$ which is possible only if $p<q$. By symmetry, we conclude that we must also have $P<Q$.

Now estimate $|a-B|\ge a-B$ and $|b-A|\ge A-b$. We obtain $$ (A+a)Pp+(a-B)pQ+(A-b)Pq+(B+b)Qq \\ =(AP+ap)+BQ(q-p)+bq(Q-P)<I\,. $$ However, the last two terms are nonnegative and the condition $0\le EX=AP-BQ$ implies that $AP\ge \frac 12[AP+BQ]=\frac 12E|X|\ge\frac 12I$ and similarly for $ap$, so we run into a contradiction.

Edit Here is the direct binomial coefficient approach. I'll start where Iosif stopped though I'll denote by $A$ what he denotes by $A/p$, etc.

We need the inequality $$ \sum_{k=0}^n{n\choose k}p^kq^{n-k}|Ak-B(n-k)|\ge Ap+Bq $$ Fix $A+B=1$ and consider the difference as a function of $A\in[0,1]$. It is piecewise linear, so the minimum is attained at a zero of some absolute value. Clearly, the endpoints (when the random variable preserves sign) are not competitors, so it is enough to consider the case when $A=\frac{n-\ell}n, B=\frac{\ell}n$, $0<\ell<n$. Then all the absolute values except the vanishing one for $k=\ell$ are at least $A+B=1$, so the LHS is at least $ 1-{n\choose \ell}p^\ell q^{n-\ell} $ and we want to show that $$ 1-{n\choose \ell}p^\ell q^{n-\ell}\ge Ap+Bq\,, $$ i.e., $$ Aq+Bp\ge {n\choose \ell} p^{\ell}q^{n-\ell} $$ Since $Aq+Bp\ge q^Ap^B=p^{\frac{\ell}{n}}q^{\frac{n-\ell}{n}}$, it suffices to check that $$ {n\choose \ell} [p^{\ell}q^{n-\ell}]^{1-\frac 1n}\le 1\,. $$ The product in brackets is maximized when $p=\frac \ell n$, $q=\frac{n-\ell}{n}$, so we want $$ {n\choose \ell}[(\tfrac{\ell}n)^{\ell}(\tfrac{n-\ell}n)^{n-\ell}]^{1-\frac 1n}\le 1\,. $$ Now the LHS can be rewritten as $\frac{n}{\ell^{\frac \ell n}(n-\ell)^{\frac{n-\ell}n}}U$ where $$ U={n-1\choose \ell-1}(\tfrac{\ell}n)^{\ell-1}(\tfrac{n-\ell}n)^{n-\ell}={n-1\choose \ell}(\tfrac{\ell}n)^{\ell}(\tfrac{n-\ell}n)^{n-\ell-1} $$ Since $U$ appears as two equal terms in the binomial expansion of $ [\tfrac{\ell}{n}+\tfrac{n-\ell}n]^{n-1}\,, $ we must have $U\le \frac 12$. On the other hand, $\frac 1{\alpha^\alpha\beta^\beta}\le 2$ for $\alpha,\beta>0, \alpha+\beta=1$, and we are done again.

It would be interesting to see what factor on the right hand side can be put in place of $1$ for $n\ge 3$. I suspect that the worst case for even $n$ is addition of $n$ Rademacher independent random variables ($\pm 1$ with probability $\frac 12$ each) even if we allow them to be different (but with expectations of the same sign) but I have no proof of it at the moment. The odd case may have a rather ugly answer even for $n=3$.

Answer 2

A more conveniently written and slightly more general version of the desired result is as follows: $$E\Big|\sum_{i\in[n]}X_i\Big|\ge E|X_1|,$$ where the $X_i$'s are iid random variables (r.v.'s) with a finite mean.

Let $p:=P(X_i\ge0)$, $q:=1-p$, $A:=EX_i^+$, $B:=EX_i^-$. Without loss of generality, $0<p<1$. For $J\subseteq[n]$, let $$I_J:=1(X_j\ge0\ \forall j\in J,X_j<0\ \forall j\notin J).$$ Following the lines of the first part of the proof for $n=2$, we have $$\begin{aligned} E\Big|\sum_{i\in[n]}X_i\Big| &=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\Big|\,I_J \\ &=\sum_{J\subseteq[n]}E\Big|\sum_{i\in[n]}X_i\, I_j\Big| \\ &=\sum_{J\subseteq[n]}E\Big|\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\, I_J\Big| \\ &\ge\sum_{J\subseteq[n]}\Big|E\Big(\sum_{i\in J}X_i^+-\sum_{i\notin J}X_i^-\Big)\, I_j\Big| \\ &=\sum_{J\subseteq[n]}\big||J|A/p-(n-|J|)B/q\big|\,p^{|J|}q^{n-|J|} \\ &=\sum_{k=0}^n\big|kA/p-(n-k)B/q\big|\,p^{k}q^{n-k}\binom nk=:s_n(p,A,B). \end{aligned}$$ Note also that $$s_n(p,A,B)=E\big|AX/p-(n-X)B/q\big|,$$ where $X$ is a r.v. with the binomial distribution with parameters $n,p$.

Without loss of generality, $A+B=1$. It remains to show that $$s_n(p,A,1-A)\ge1$$ for all $A\in[0,1]$. Graphing suggests this is true.

Unfortunately, the second, "squaring" part of the proof for $n=2$ will not work even for $n=3$ and $p$ close enough to $1/2$; that is, the inequality $$\sum_{k=0}^n\Big(\big(kA/p-(n-k)B/q\big)\,p^{k}q^{n-k}\binom nk\Big)^2\ge(A+B)^2$$ will not hold for such $n,p$.