This question is concerning the paper, particularly the proof of Lemma 2.1 in Section 2.1:


* Matas, A., Merker, J. *Existence of weak solutions to doubly degenerate diffusion equations*, Appl Math **57** (2012) pp43–69, doi:[10.1007/s10492-012-0004-0](https://doi.org/10.1007/s10492-012-0004-0), [EuDML](https://eudml.org/doc/246610).


Let $m \geq 2$ and let $m'$ be its conjugate. Let $w\_j$ for $j=1, ..., k$ be a basis of $H\_1 \cap L^{m'}$. The task is to show that there is a $u(t) \in \text{span}(w\_1, ..., w\_k)=:A$ such that
$$\langle\varphi(u(t)), w\_j \rangle\_{L^m, L^{m'}} = \langle \varphi(u\_0), w\_j \rangle\_{L^m, L^{m'}} + \int\_0^t \langle F(s, u(s)), w\_j \rangle\mathrm{d}s$$
for all $j=1, ..., k$.


Here, $\varphi:L^{m'} \to L^m$ is $\varphi(u) = u^{m'-1}$ and
$$\langle F(s,u), w\_j \rangle = -\langle \nabla u, \nabla w\_j \rangle.$$
In order to show existence, the author shows that $\langle F(s,u), w\_j \rangle$ is locally Lipschitz continuous on the finite-dimensional subspace:
$$|\langle F(s,u), w\_j \rangle -\langle F(s,v), w\_j \rangle| \leq C|u-v|\_{H^1}$$ with $C$ depending on $u\_0$.


And author states that "$\varphi$ has a locally Lipschitz continuous inverse" but the author only seems to prove
$$|u-v|\_A \leq C|\varphi(u) - \varphi(v)|\_{L^m}.$$


Then it is stated



> 
> The integral equation has a solution due to Banach's fixed point theorem in analogy to the existence theorem of Picard-Lindelof.
> 
> 
> 


From an answer to [a thread I posted to MSE](https://math.stackexchange.com/questions/794563/banach-fixed-point-theorem-and-picard-lindelof-applied-to-this-equation-explana), it is suggested that we consider
$$\langle y\_{n+1}(t), w\_j \rangle = \langle \varphi(u\_0), w\_j \rangle + \int\_0^t \langle F(s, \varphi^{-1}(y\_n)(s)), w\_j \rangle\mathrm{d}s$$
But then $y\_{n+1}$ need not be finite-dimensional and so I don't see how to use the Lipschitz condition on $\varphi^{-1}$. Even if we assume $y\_{n}$ are finite-dimensional, why should $\varphi^{-1}(y\_n) =: u\_n$ be finite-dimensional as required? I don't know what the right way to do this is. Any help appreciated.


(I chose $p=2$ and the right hand side $f\equiv 0$).



Existence of an integral equation (Faedo-Galerkin, Banach fixed point, Picard-Lindelof)
OK, I think I figured it out.$\newcommand\esssup{\operatorname{ess} \sup}$


First let me define a few things: for any given basis $\{w\_1\ldots w\_k\}$ of $A\subset L^{m'}$ let $\{w'\_1,\ldots w'\_k\}$ be the canonical dual basis of $A'$ (the finite-dimensional dual), i-e $\langle w'\_i,w\_j\rangle\_{A',A}=\delta\_{ij}$. In particular the $w'\_i$'s can be seen as functions $w'\_i(x)\in L^m$. Thus any linear form $v$ on $A$ can be written as $v(x)=\sum\limits\_{i=1}^k\lambda\_iw'\_i(x)\in A'$, and for all $u(x)=\sum\limits\_{i=1}^ky\_iw\_i(x)\in A$ we can compute the action
$$
\langle v,u\rangle\_{A',A}= \langle v,u\rangle\_{L^m,L^{m'}}=\int\_{\Omega}v(x)u(x)\,dx
$$
by $L^m$ duality. For any $v\in L^m$ (and not necessarily in $A'$, this is important) we can moreover define the "projection" $P\_A$ onto $A'$ by
$$
\forall u=\sum\_1^ky\_i w\_i\in A:\qquad \langle P\_A v,u\rangle\_{A',A}=\langle v,u\rangle\_{L^m,L^{m'}}=\int v(x)u(x)\,dx.
$$
Let also $\varphi\_A=P\_A\circ \varphi:A\to A'\subset L^m$. Now for given coefficients $\lambda\_1,\ldots,\lambda\_k$ the equation
$$
\forall j=1\ldots k:\qquad \langle\varphi(u),w\_j\rangle\_{L^m,L^{m'}}=\lambda\_j\hspace{2cm}(1)
$$
actually reads
$$
\text{find }u\in A\text{ such that }\varphi\_A(u)=v\text{ in }A',
$$
where $v(x)=\sum\_1^k\lambda\_j w'\_j(x)$ can be unambiguously considered either as a function in $L^m(\Omega)$ or as an element of $A'$.


**Claim 1**: given any $\lambda=\lambda\_1,\ldots,\lambda\_k$ equation (1) admits a unique solution $u(x)=\sum\_{i=1}^k y\_iw\_i(x)\in A$. This comes from quite simple finite dimensional considerations by looking at the restriction $\Phi\_A$ of $\Phi(u)=|u|\_{L^{m'}}^{m'}$ to the subspace $A$, and noting that for all $u\in A$ the differential can be computed as
$$
\forall\, h \in A:\qquad \langle D\Phi\_A(u),h\rangle\_{A',A}= \langle\varphi(u),h\rangle\_{L^m,L^{m'}}= \int \varphi(u)(x)h(x)\,dx.
$$
Please let me know if you need details (roughly, $\varphi\_A$ is the gradient of a $\mathcal{C}^1(A)$ strictly convex and superlinear function $\Phi\_A$, so in particular it is a bijection). This claim 1 defines in particular an inverse $\psi\_A:=\varphi\_A^{-1}:A'\to A$. This is just a fancy way to say that, when viewed as taking values in the dual $A'$ (by restricting the $L^m$ action to test functions in $A\subset L^{m'}$), the map $\varphi$ restricted to $A$ is invertible. This should come as no surprise since the map $\varphi(z)=z^{m'-1}$ is increasing, which is exactly what makes the PDE formally parabolic. Compared to the paper [Alt-Luckhaus] I already mentioned, this corresponds to the fact that $\varphi$ is the gradient of the convex function $\Phi$, hence a monotone vector-field (easier here since the setting is scalar, but this is really what it is).


**Claim 2:** the inverse $\psi\_A$ form step 1 is locally Lipschitz. This is just a refinement of your paper, but with an extra subtlety. Let $v^1=\sum \_{i=1}^k\lambda\_i^1w'\_i$ and $v^2=\sum\_{i=1}^k \lambda\_i^2w'\_i$ be two given elements of $A'$, and denote the corresponding solutions to (1) by $u^1=\psi\_A(v^1)$ and $u^2=\psi\_A(v^2)$. Using the convexity inequality
$$
|a-b|^2\leq C(a^{m'-1}-b^{m'-1})(a-b)(|a|+|b|)^{2-m'}
$$
you get
\begin{align\*}
|u^1-u^2|^2\_{L^2} & \leq C\int\_{\Omega}((u^1)^{m'-1}-(u^2)^{m'-1})(u^1-u^2)(|u^1|+|u^2|)^{2-m'}\,dx\\
 & \leq C(\esssup |u^1|+\esssup |u^2|)^{2-m'}\int\_{\Omega}((u^1)^{m'-1}-(u^2)^{m'-1})(u^1-u^2)\,dx.
\end{align\*}
In the last line I used the fact that $m'>1$ (since $m<\infty$) and monotonicity of $z\mapsto z^{m'-1}$ so that the last integrand is really pointwise nonnegative a.e $x\in \Omega$. Now because $u^1,u^2$ actually lie in the finite dimensional subspace $A$, you can rewrite this as
\begin{align\*}
|u^1-u^2|^2\_{L^2} & \leq C(\esssup |u^1|+\esssup |u^2|)^{2-m'}\int\_{\Omega}(\varphi(u^1)-\varphi(u^2))(u^1-u^2)\,dx\\
& =C(\esssup |u^1|+\esssup |u^2|)^{2-m'}\int\_{\Omega}(\varphi\_A(u^1)-\varphi\_A(u^2))(u^1-u^2)\,dx.
\end{align\*}
Indeed the $\varphi(u^1),\varphi(u^2)\in L^m$ terms above "see" test functions $u^1,u^2\in A$, so their action is the same as that of their projection $\varphi\_A(u^1),\varphi\_A(u^2)$. Applying Young's inequality you get
$$
|u^1-u^2|^2\_{L^2} \leq C(\esssup |u^1|+\esssup |u^2|)^{2-m'}|\varphi\_A(u^1)-\varphi\_A(u^2)|\_{L^m}|u^1-u^2|\_{L^{m'}},
$$
and by equivalence of the norms on $A$ you conclude that
$$
\forall \,u^1,u^2\in A:\qquad |u^1-u^2|\_A\leq C(\esssup |u^1|+ \esssup |u^2|)^{2-m'}\varphi\_A(u^1)-\varphi\_A(u^2)|\_{L^m}|,
$$
which really means that the inverse $\psi\_A=\varphi\_A^{-1}$ is locally Lipschitz continuous if $u^1,u^2$ stay in a ball around some $u\_0$. The trick was here essentially to control $|u^1-u^2|$ not by $|\varphi(u^1)-\varphi(u^)|$, which a priori takes values in the "large" $L^m$ space, but rather by their projection $|\varphi\_A(u^1)-\varphi\_A(u^2)|$ onto the dual $A'$, which is finite-dimensional ("small"). I guess this is what the authors had in mind in their paper, but it was well hidden and quite misleading.


**Step 3: back to the fixed point in the integral formulation.**
$$
\forall j=1\ldots k:\qquad \langle\varphi(u(t)), w\_j \rangle\_{L^m, L^{m'}} = \langle \varphi(u\_0), w\_j \rangle\_{L^m, L^{m'}} + \int\_0^t \langle F(s, u(s)), w\_j \rangle\mathrm{d}s
$$
By the classical Lipschitz estimates in the OP the right hand-side defines, for given $u\in C(0,T;A)$ a Lipschitz continuous map $u\mapsto V[u]\in C(0,T;A')$. In other words, the above integral formulation reads:
$$
\text{find }u\in C(0,T;A)\text{ such that}\qquad \phi\_a[u]=V[u]\text{ in }C(0,T;A').
$$
Now because $\psi\_A=\phi\_A^{-1}$ is locally Lipschitz continuous you can easily rewrite the fixed point as
$$
\text{find }u\in C(0,T;A)\text{ such that}\qquad u=\psi\_A\circ V[u]\text{ in }C(0,T;A).
$$
Finally tuning the maximal time $T$ (small) and initial condition (stay in a given large ball around the "initial datum" to take care of the $\esssup$'s) it should be straightforward to prove that $\psi\_A\circ V$ is a contraction in $C(0,T;A)$, hence the desired projected solution.


