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 $L^p(\Omega \times \Omega)$.
 A: 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}]

A: 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 $2^{-1/p}$. 
