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Let $(\Omega,\mu)$ be a finite measure space such that $\mu(\Omega)=1$. Suppose $1\leq p \leq \infty$.

Let $\psi \colon L^p(\Omega) \to L^p(\Omega \times \Omega)$ be the map which maps $f$ onto the function $(x,y)\mapsto \frac{1}{2}\big(f(x)+f(y)\big)$. The map $\psi$ is contractive.

1) Is it an isomorphic embedding?

The answer is positive (see below).

Follow-up questions:

2) What is the best constant $c$ in $\|\psi(f)\|_p\ge c\|f\|_p$?

3) Does there exist a bounded projection from $L^p(\Omega \times \Omega)$ onto the range of $\psi$? Remark: the answer to Question 1 imply that the range of $\psi$ is a closed subspace of $L^p(\Omega \times \Omega)$.

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    $f(x)f(y)$ looks very natural, although it is not linear.2012-08-14

2 Answers 2

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Claim: $\|\psi(f)\|\ge \dfrac15\|f\|_p$ for all $1\le p\le \infty$. (Optimized by @timur).

Proof. The proof applies to either real or complex-valued functions, but I'm going to assume they are real. Consider two cases.

(a) there exists $c\in\mathbb R$ such that $\|f-c\|_p<\dfrac{2}{5} \|f\|_p$. By the triangle inequality $|c|\ge \dfrac{3}{5}\|f\|_p$. The linearity and contractivity of $\psi$ imply $\|\psi(f)\|_p =\|\psi(c)+\psi(f-c)\|_p \ge \|\psi(c)\|_p - \|\psi(f-c)\|_p \ge |c|-\frac{2}{5} \|f\|_p \ge \frac15 \|f\|_p.$

(b) for all $c\in\mathbb R$ we have $\|f-c\|_p\ge \dfrac{2}{5} \|f\|_p$. Then for every $y\in \Omega$ $\left\|\frac12(f(\cdot)+f(y))\right\|_p\,dx \ge \frac{1}{5} \|f\|_p$ Raising to power $p$ and integrating over $y$ (or, if $p=\infty$, applying the definition of the $L^\infty$ norm), we obtain $\|\psi(f)\|_p\ge \dfrac{1}{5}\|f\|_p$. $\quad\Box$

Follow-up question. What is the best constant $c$ in $\|\psi(f)\|_p\ge c\|f\|_p$? The proof gives $1/5$ and the example $f(x)=\chi_{[0,1/2]}-\chi_{[1/2,1]}$ shows we can't get more than $2^{-1/p}$.

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    Various plots of the function (for different values of $p$) suggest this to be true. The symmetry of the (general) inequality indicates, that there may also be a more direct proof using some techniques related to Shur convexity.2012-08-17
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There was asking to prove that the best coercitivity constant $c_p$ for $\psi$ is $2^{-1/p}$. In fact this is not true.

For a given simple function $ f=\sum\limits_{k=1}^n a_k\chi_{A_k} $ denote $x_k=\mu(A_k)$. Consider special case $a_1=-1$, $a_2=0$, $a_3=1$ and $x_1=\varepsilon$, $x_2=1-2\varepsilon$, $x_3=\varepsilon$ where $\varepsilon\in(0,2^{-1})$. Then $ c_p\leq\Vert\psi(f)\Vert_p/\Vert f\Vert_p=(\varepsilon+2^{1-p}(1-2\varepsilon))^{1/p} $ Since left hand side is independent of $\varepsilon$ we conclude $ c_p\leq\min_{\varepsilon\in(0,2^{-1})}(\varepsilon+2^{1-p}(1-2\varepsilon))^{1/p}=2^{(1-\max(2,p))/p} $ But even the bound $ b_p=2^{(1-\max(2,p))/p} $ is not rough. Numeric test showed that for $p=3$, $a_1=0.079$, $a_2=0.079$, $a_3=-1$ with $x_1=0.879$, $x_2=0.99$, $x_3=0.022$ gives $ c_3< 0.612176<0.629960\approx b_3 $

Here is a Mathematica code to check this

FNorm[a_, x_, n_, p_] := (Sum[Abs[a[[k]]]^p x[[k]], {k, 1, n}])^(1/p); FImageNorm[a_, x_, n_,     p_] := (Sum[      Abs[(a[[k]] + a[[l]])/2]^p x[[k]] x[[l]], {l, 1, n}, {k, 1,        n}])^(1/p); FOpNorm[a_, x_, n_, p_] := FImageNorm[a, x, n, p]/FNorm[a, x, n, p]  OpNorm = 1; A = {}; X = {}; p = 3; With[{n = 3, R = 1, M = 100000}, For[i = 0, i < M, i++,   a = RandomReal[{-R, R}, n];   x = RandomVariate[GammaDistribution[1, 1], n];   x = x/Total[x];   norm = FOpNorm[a, x, n, p];   If[norm < OpNorm, {OpNorm, A, X} = {norm, a, x}, Continue[]];   ]  ] Print[{{OpNorm, 2.^((1-Max[2,p])/p)}, A, X}] 
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    @LVK Thanks for your bounty! As for the question. I don't think sets $A_1$ and $A_2$ can't be combined because values of $f$ on this sets are different. My computational research that the bound $b_p$ seems to be correct for $p\in[1,2]$, but wrong for other values. It seems to me that one need to restrict its research to the case $\Omega=[0,1]$. My intuition tells me that we need to consider Chebyshev polynomials...2012-09-30