# Isomorphic embedding of $L^{p}(\Omega)$ into $L^{p}(\Omega \times \Omega)$?

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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Thanks for the comments. If $p=\infty$, the map $\psi$ is really an isometry. – Zouba Aug 8 '12 at 14:43
If $1 \leq p <\infty$, $\psi$ is definitely not an isometry but I think (hope) that $\psi$ is an isomorphic embedding. – Zouba Aug 8 '12 at 14:45
It's injective, hence an isomorphims on its range. – Davide Giraudo Aug 8 '12 at 14:50
@DavideGiraudo, but we mus say thay range of $\psi$ is closed to be sure that $\psi$ is an isomorphism on its range. – Norbert Aug 8 '12 at 21:14
@Zouba No, we have $\psi(f)\ge 1$ on the rectangle $[0,\epsilon]\times [0,1]$, so its $L^p$ norm is of size $\epsilon^{1/p}$. More generally, if $f\ge 0$ on $\Omega$, then $\psi_f(x,y)=\frac12 ( f(x)+f(y))\ge \frac12 f(x)$, hence $\|\psi(f)\|_p\ge\frac12 \|f\|_p$. So the only potential issue is when $f$ has both positive and negative values, creating potential for cancellation. – user31373 Aug 11 '12 at 19:26

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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Replacing $\frac15$ in $\|f-c\|_p<\frac15\|f\|_p$ by $0<\gamma<1$ and optimizing gives $c=\frac15$ in $\|\psi(f)\|_p\geq c\|f\|_p$ (with $\gamma=\frac25$). – timur Aug 14 '12 at 1:52
@timur Thanks, but I don't expect this (a)-(b) split to ever give the sharp constant. As far as examples go, I can't do better than $f(x)=\chi_{[0,1/2]}-\chi_{[1/2,1]}$ for which $c=2^{-1/p}$. – user31373 Aug 14 '12 at 2:02
Now I am a bit confused: With $f$ as above, isn't it $\|f\|_1=\frac12$ and $\|\psi(f)\|_1=\frac13$, hence $c=\frac23$? – timur Aug 14 '12 at 3:16
@Zouba I'm sure it's been used before, but can't tell you where. Concerning your 'addon': I guess there is a projection, simply because you shouldn't get explicit uncomplemented subspaces of $L^p$ this easily. Of course there is a canonical projection when $p=2$, and maybe by writing it down explicitly you'll see a way to generalize it. If you get stuck, I recommend asking this as a separate question (with a link here), because not many people will see your question buried in the comments. – user31373 Aug 16 '12 at 16:38
I have made some progress towards a proof of the optimal lower bound conjectured by timur. Unfortunately I do not have the time right now to work out the details, but I want to share the idea. Using the density of the subspace generated by characteristic functions of measurable subsets, it is possible to reduce the problem to showing that the discrete inequality $\sum_{i=1}^n\sum_{j=1}^n \vert a_i + a_j \vert^p x_ix_j \geq 2^{p-1}$ holds for all $x_1, \dots, x_n \geq 0$ and $a_1, \dots, a_n \in \mathbb R$ with $x_1 + \dots + x_n = 1$ and $\vert a_1 \vert^px_1 + \dots + \vert a_n \vert^px_n=1$. – Alexander Thumm Aug 17 '12 at 9:51

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];

Interesting indeed. Actually, the terms of the bounty were "to prove or disprove.." so your answer qualifies. I guess the natural from here is to understand what's the worst that can happen for $p=1$. – user31373 Sep 15 '12 at 20:21
@LVK So your question is what is the norm of $\psi$ when $p=1$? – Norbert Sep 16 '12 at 4:26
I'm just thinking this would be the easiest case to understand. Do you have an explicit example with constant less than $1/2$ for $p=1$? – user31373 Sep 16 '12 at 4:46
My current guess for the best constant: $2^{-1/p}$ when $p\le 2$ and $2^{(1-p)/p}$ for $p\ge 2$. The last bit is actually odd because it tends to $1/2$ as $p\to \infty$, despite the map being an isometry when $p=\infty$. Details here – user31373 Sep 26 '12 at 4:10