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Let $M$ be a compact and connected manifold without boundary. My question is how to prove the following fact which I believe is true:

If $f : M \to \mathbb{R}$ is a continuous function that attains the values $a < b$, then for any $c\in [a,b]$ and any $1\leq p<\infty$, there is diffeomorphism $\varphi : M\to M$ such that $f \circ \varphi$ is close to $c$ in the $L^p(M)$ norm.

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    $\begingroup$ In this form the result surely is false. Take $f \equiv -1/2$. $\endgroup$ Commented Jan 24, 2021 at 8:12
  • $\begingroup$ Do tou know the precise statement? $\endgroup$ Commented Jan 24, 2021 at 14:44
  • $\begingroup$ O think that maybe os crutial to asume that $f$ is not constant $\endgroup$ Commented Jan 24, 2021 at 14:45
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    $\begingroup$ Please correct the statement since the first part is still not correct. If $f>-1$ on $M$, then $f\circ\phi$ will never be close to $-1$. The last statement suggested by me is correct so rewrite the question in a way that it is correct. I am working on my answer, but the question should be correctly formulated. $\endgroup$ Commented Jan 24, 2021 at 22:48
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    $\begingroup$ I edited your question. I hope you do not mind. $\endgroup$ Commented Jan 24, 2021 at 23:18

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The answer to the last question follows from the following result:

Theorem. If $f:\mathcal{M}\to\mathbb{R}$ is a continuous function on a smooth compact connected manifold without boundary and if $$ \inf_{\mathcal{M}} f\leq c\leq \sup_{\mathcal{M}} f, $$ then for any $1\leq p<\infty$ and any $\varepsilon>0$ there is a diffeomorphism $\varphi:\mathcal{M}\to\mathcal{M}$ such that $$ \Vert f\circ\varphi -c\Vert_p<\varepsilon. $$

Proof. It follows from the existence of a triangulation of $\mathcal{M}$ that there is a smooth mapping $$ p:\overline{B}^n\to\mathcal{M} $$ defined on the closed unit ball such that $p$ is a diffeomorphism in the interior $B^n$ of $\overline{B}^n$ and that $p(\partial B^n)$ is contained in the $(n-1)$-dimensional skeleton of $\mathcal{M}$. We obtain $p$ by expanding one cell of a trianglation of $\mathcal{M}$ by pushing adjacent cells to their boundaries.

Clearly there is $x_0$ such that $f(x_0)=c$. We can assume that $x_0$ is not in the $(n-1)$-skeleton of $\mathcal{M}$ so that $x_0=p(y_0)$ for some $y_0\in B^n$.

Fix $\eta>0$. Let $X$ be a smooth vector field in $B^n$ radially emerging from $y_0$ and vanishing in a neighborhood $U$ of $\partial B^n$ that has measure $|U|<\eta/2$. Let $\Phi_t$ be the flow generated by $X$ so $\Phi_t=\operatorname{id}$ in $U$. Then for any $\delta>0$, $$ |B^n\setminus\Phi_t(B^n(y_0,\delta))|<\eta $$ provided $t$ is sufficiently large. Indeed, $B^n\setminus B^n(y_0,\delta)$ will be pushed towards $U$ that has measure less than $\eta/2$.

Now we push forward the family of diffeomorphisms to $\mathcal{M}$ by the formula $$ \Psi_t(x)= \begin{cases} p(\Phi_t(p^{-1}(x))), & x\in p(B^n),\\ x & x\in p(\partial B^n). \end{cases} $$ This is a well defined family of diffeomorphisms that is identity in $p(U)$ which is a small neighborhood of the $(n-1)$-dimensional set $p(B^n)$.

The diffeomorphisms $\Psi_t$ expand tiny neighborhoods of $x_0$ towards the small set $p(U)$. Thus for $t$ sufficiently large, the function $f\circ \Psi_t^{-1}$ is very close in the uniform norm to $f(x_0)=c$ except on a set of a very small measure (a tiny neighborhood of $p(\partial B^n)$ and we can guarantee by taking small $\eta$ and large $t$ that $$ \Vert f\circ\Phi_t^{-1}-c\Vert_p<\varepsilon. $$

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