Miscellaneous TeX; changed title to be hopefully more descriptive

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edited Feb 5, 2021 at 22:57

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

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

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.

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

ProofProof by Kent Merryfield:

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

Then \begin{align*}\int_0^1 \int_0^1 |f(x) + f(y)|\,dx\,dy &= \int_P \int_P |f(x) + f(y)|\,dx,dy + \int_P \int_N |f(x) + f(y)| \,dx\, dy \\ &+ \int_N \int_P |f(x) + f(y)|\, dx \,dy+ \int_N \int_N |f(x) + f(y)|\, dx\, dy\end{align*}\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 |f(x) + f(y)|\, dx\, dy &= \int_P \int_P (f(x) + f(y))\,dx \,dy\\ &= |P| \int_P f(x)\,dx + |P| \int_P f(y) \,dy = 2|P| \int_P f^+(x) \,dx\end{align*}\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 $|P|$$\abs P$ to mean the measure (total net length) of the set $P.$$P$.

Similarly, $\int_N \int_N |f(x) + f(y)| \,dx \,dy = 2|N| \int_N f^-(x) \,dx.$$\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 |f(x) + f(y)| \,dx \,dy &= \int_P \int_N |f^+(x) - f^-(y)| \,dx \,dy\\ &\ge \left|\int_P \int_N f^+(x) - f^-(y) \,dx \,dy\right|\\ &= \left| |N| \int_P f^+(x) \,dx - |P| \int_N f^-(y) \,dy \right|\end{align*}\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,$$A = \int_P f^+(x) \,dx$, $B = \int_N f^-(x) \,dx,$$B = \int_N f^-(x) \,dx$, and $I = \int_0^1 \int_0^1 |f(x) + f(y)| \,dx \,dy,$$I = \int_0^1 \int_0^1 \abs{f(x) + f(y)} \,dx \,dy$, then we have found that $I \ge 2|P|A + 2|N|B + 2|(|N|A - |P|B)|.$$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[(|P|A + |N|B)^2 + (|N|A - |P|B)^2 + (\text{other positive terms})\right]\\ &\ge 4(|P|^2A^2 + |N|^2B^2 + |N|^2A^2 + |P|^2B^2)\\ &= 4(|P|^2 + |N|^2)(A^2 + B^2).\end{align*}\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,$$b$, $(a + b)^2 \le 2(a^2 + b^2)$ since $2(a^2 + b^2) - (a + b)^2 = (a - b)^2.$$2(a^2 + b^2) - (a + b)^2 = (a - b)^2$.

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

Is $\iiint |f(x)+f(y)+f(z)|\, dx\, dy\, dz \ge \int|f(x)|\, dx$ on $[0,1]$?

Question: Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Does the following inequality hold? $$\int_{0}^{1}\int_{0}^{1}\cdots\int_0^1\int_0^1|f(x_{1})+f(x_{2})+\cdots+f(x_{n})|dx_1 \; dx_{2}\cdots;dx_{n} \ge \int_0^1 |f(x)|dx$$

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 $$\int_0^1\int_0^1|f(x)+f(y)|dx \; dy \ge \int_0^1 |f(x)|dx$$

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

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

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

For convenience, we now square this: \begin{align*}I^2 &\ge 4\left[(|P|A + |N|B)^2 + (|N|A - |P|B)^2 + (\text{other positive terms})\right]\\ &\ge 4(|P|^2A^2 + |N|^2B^2 + |N|^2A^2 + |P|^2B^2)\\ &= 4(|P|^2 + |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, $2(|P|^2 + |N|^2) \ge (|P| + |N|)^2 = 1^2,$ since it is the measure of the interval $[0, 1].$ Also, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 |f(x)| \,dx\right)^2.\ \square$

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

$\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Abs[1]{\left\lvert#1\right\rvert}$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, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 \abs{f(x)} \,dx\right)^2$. $\square$

put the question at the top

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edited Feb 5, 2021 at 9:52

user44143

prove this integeral inequality with putnam 2003 probelm Is $\iiint |f(x)+f(y)+f(z)|\, dx\, dy\, dz \ge \int|f(x)|\, dx$ on $[0,1]$?

Question: Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Does the following inequality hold? $$\int_{0}^{1}\int_{0}^{1}\cdots\int_0^1\int_0^1|f(x_{1})+f(x_{2})+\cdots+f(x_{n})|dx_1 \; dx_{2}\cdots;dx_{n} \ge \int_0^1 |f(x)|dx$$

Even the case $n=3$ would be interesting. This is inspired by an oldthe case $n=2$, which was a Putnam problem from 2003, and a solution given by composite of solutions by Kent Merryfield:as follows.

Question (Putnam 2003)LetTheorem. Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Show thatThen $$\int_0^1\int_0^1|f(x)+f(y)|dx \; dy \ge \int_0^1 |f(x)|dx$$ Kent Merryfield's proof

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

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

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

For convenience, we now square this: \begin{align*}I^2 &\ge 4\left[(|P|A + |N|B)^2 + (|N|A - |P|B)^2 + (\text{other positive terms})\right]\\ &\ge 4(|P|^2A^2 + |N|^2B^2 + |N|^2A^2 + |P|^2B^2)\\ &= 4(|P|^2 + |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, $2(|P|^2 + |N|^2) \ge (|P| + |N|)^2 = 1^2,$ since it is the measure of the interval $[0, 1].$ Also, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 |f(x)| \,dx\right)^2 .$

MY Question 1:Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Show that $$\int_{0}^{1}\int_0^1\int_0^1|f(x)+f(y)+f(z)|dx \; dy\;dz \ge \int_0^1 |f(x)|dx$$ and I think maybe this generality also is hold:

MY Question 2:Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Show that $$\int_{0}^{1}\int_{0}^{1}\cdots\int_0^1\int_0^1|f(x_{1})+f(x_{2})+\cdots+f(x_{n})|dx \; dx_{2}\cdots;dx_{n} \ge \int_0^1 |f(x)|dx$$ and Now How to prove.Thanks$2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 |f(x)| \,dx\right)^2.\ \square$

prove this integeral inequality with putnam 2003 probelm

This is inspired by an old Putnam problem from 2003, and a solution given by composite of solutions by Kent Merryfield:

Question (Putnam 2003)Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Show that $$\int_0^1\int_0^1|f(x)+f(y)|dx \; dy \ge \int_0^1 |f(x)|dx$$ Kent Merryfield's proof:

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

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

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

For convenience, we now square this: \begin{align*}I^2 &\ge 4\left[(|P|A + |N|B)^2 + (|N|A - |P|B)^2 + (\text{other positive terms})\right]\\ &\ge 4(|P|^2A^2 + |N|^2B^2 + |N|^2A^2 + |P|^2B^2)\\ &= 4(|P|^2 + |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, $2(|P|^2 + |N|^2) \ge (|P| + |N|)^2 = 1^2,$ since it is the measure of the interval $[0, 1].$ Also, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 |f(x)| \,dx\right)^2 .$

MY Question 1:Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Show that $$\int_{0}^{1}\int_0^1\int_0^1|f(x)+f(y)+f(z)|dx \; dy\;dz \ge \int_0^1 |f(x)|dx$$ and I think maybe this generality also is hold:

MY Question 2:Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Show that $$\int_{0}^{1}\int_{0}^{1}\cdots\int_0^1\int_0^1|f(x_{1})+f(x_{2})+\cdots+f(x_{n})|dx \; dx_{2}\cdots;dx_{n} \ge \int_0^1 |f(x)|dx$$ and Now How to prove.Thanks

Is $\iiint |f(x)+f(y)+f(z)|\, dx\, dy\, dz \ge \int|f(x)|\, dx$ on $[0,1]$?

Question: Let $f(x)$ be a continuous real-valued function deﬁned on the interval $[0, 1]$. Does the following inequality hold? $$\int_{0}^{1}\int_{0}^{1}\cdots\int_0^1\int_0^1|f(x_{1})+f(x_{2})+\cdots+f(x_{n})|dx_1 \; dx_{2}\cdots;dx_{n} \ge \int_0^1 |f(x)|dx$$

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 $$\int_0^1\int_0^1|f(x)+f(y)|dx \; dy \ge \int_0^1 |f(x)|dx$$

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

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

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

For convenience, we now square this: \begin{align*}I^2 &\ge 4\left[(|P|A + |N|B)^2 + (|N|A - |P|B)^2 + (\text{other positive terms})\right]\\ &\ge 4(|P|^2A^2 + |N|^2B^2 + |N|^2A^2 + |P|^2B^2)\\ &= 4(|P|^2 + |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, $2(|P|^2 + |N|^2) \ge (|P| + |N|)^2 = 1^2,$ since it is the measure of the interval $[0, 1].$ Also, $2(A^2 + B^2) \ge (A + B)^2 = \left(\int_0^1 |f(x)| \,dx\right)^2.\ \square$

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asked Feb 5, 2021 at 6:28

math110

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