I want to prove a theorem which I urgently need, but i am kind of stuck at a certain point. My Problem is to prove the following: EDIT I found a proof, but I'm not sure if there are any mistakes in it. Any help appreciated, also if you find a different proof.
Let $\Omega\subset\mathbb{R}^n$ be an open, bounded subset and $p\in\left(1,\infty\right)$ such that $L^p\left(\Omega\right)$ is reflexive.
For $f\in C^{1}\left(\Omega\right)$ let $f_{n}\in C^{1}\left(\Omega\right)$ be a sequence such that $f_n\rightrightarrows f$ and $\nabla f_n\rightrightarrows \nabla f$ are uniformly convergent. EDIT $f$ and $f_n$ shall be non negative. $\nabla f_n$ and $\nabla f$ are not used.
Furthermore let $g_n\in L^{p}\left(\Omega\right)$ be simple functions such that $g_n\stackrel{L^p}{\rightharpoonup} g\in L^p\left(\Omega\right)$. (See EDIT2 for what I mean by simple functions here.)
Let $D\subset\Omega$ be a dense subset such that $\forall x\in D \exists n_0\in\mathbb{N}\forall n\geq n_0:g_n\left(x\right)<\infty$ and $g_n\left(x\right)=\left|\ln\left(f_{n}\left(x\right)\right)\right|.$ EDIT I dont need $D$ anymore. Instead let $g_{n}\left(x_{n,i}\right)$ be always finite with $g_{n}\left(x_{n,i}\right)=\left|\ln\left(f_{n}\left(x_{n,i}\right)\right)\right|$.
Now I want to prove that $g=\left|\ln\left(f\right)\right|$ almost everywhere.
Thanks a lot in advance for your time!
EDIT1:(Replaced by EDIT2)
I call a function simple, when it equals a finite linear combinations of Characteristic functions for cubes of positive measure. (products of intervals instead of arbitrary measurable sets) There is a bit of freedom here since the simple functions stem from a quadrature-rule.
EDIT2:
Let $g_n$ be simple functions inspired by Riemann-Integration:
More precisely, let $g_n=\sum_{j=1,...,N_n} g\left(x_{j,n}\right)\chi_{Q_{j,n}}$ be such that $x_{j,n}\in Q_{j,n}$ with $g\left(x_{j,n}\right)<\infty$ and $Q_{j,n}$ are products of intervals (of positive measure) such that for $i\neq j$ holds: $\mathrm{int}\left(Q_{i,n}\right)\cap \mathrm{int}\left(Q_{j,n}\right)$ and $\lim_{n\in\mathbb{N}}\max_{i=1,...,N_n}\left(\mathrm{diam}\left(Q_{i,n}\right)\right)=0.$
Also let $f, f_n$ be non-negative.
remarks:
- The problem here is that $\ln$ is discontinuous at $0$.
- If there would be a subsequence of $g_n$ converging almost everywhere then the rest would be easy. (But i dont find any reason for this to exist.)
- if nescessairy for $\Omega$ more restrictions are possible
- also for $\nabla\ln f_n$ there exists a sequence $h_n\rightharpoonup h$ of simple functions such that $\forall x\in D\exists n_0\in\mathbb{N}\forall n\geq n_0:h_n\left(x\right)<\infty,h_n\left(x\right)=\nabla\ln\left(f_{n}\left(x\right)\right)$
- $f_n\rightarrow f$ also in the sense of $C^{1,\alpha}$ for an $\alpha>0$ if nescessairy
- The simple functions occur because of a quadrature-formula that I use, so I cannot really replace them by something else. (At least I have no idea what that could be) but I have a bit of freedom of choice.
Answering my own question: (Could you check my proof please? That would help me a lot. I need to know if I am wrong or not.)
Since $f$ and $f_n$ are continuous and bounded they are Riemann integrable and by uniform convergence $f_n\rightrightarrows f$ it holds that $\sum_{i=1,...,N_n}f_n\left(x_{i,n}\right)\chi_{Q_{i,n}}\stackrel{L^1}{\rightarrow}f.$ (also almost everywhere, even everywhere.)
Now assume $\lambda\left(f^{-1}\left\{0\right\}\right)\geq\epsilon$ for an $\epsilon> 0.$ Then there is a test function $\psi\in C_{0}^{\infty}$ with the following properties:
$\mathrm{img}\left(\psi\right)\subset\left[0,1\right]$
$\lambda\left(\psi^{-1}\left(\left\{0\right\}\right)\right)\leq\frac{\epsilon}{3}$
$\lambda\left(\psi^{-1}\left(\left(0,1\right)\right)\right)\leq\frac{\epsilon}{3}$
By Fatou and weak convergence we have: \begin{equation} \infty>\int\psi\cdot g = \lim_{n\in\mathbb{N}}\int\psi\cdot g_n\geq \int \liminf_{n\in\mathbb{N}}\psi\cdot g_n. \end{equation} Let $x\in f^{-1}\left(\left\{0\right\}\right).$ Then (since $f_{n}\left(x\right)< 1$ and thus $\left|\ln\left(f_{n}\left(x\right)\right)\right|=-\ln\left(f_{n}\left(x\right)\right)$ for $n$ sufficiently large) it holds $e^{-g_{n}\left(x\right)}=f_{n}\left(x\right)\rightarrow 0$ and therefore $g_n\left(x\right)\rightarrow +\infty$ meaning $\forall M\geq 0\exists n_{0}:\forall n\geq n_0: g_n\left(x\right)\geq M.$ By monotonicity of $\int$ it holds: \begin{equation} \int \liminf_{n\in\mathbb{N}}\psi\cdot g_n\geq\int_{f^{-1}\left(\left\{0\right\}\right)\cap\psi^{-1}\left(\left\{1\right\}\right)} \liminf_{n\in\mathbb{N}}\psi\cdot g_n= \int_{f^{-1}\left(\left\{0\right\}\right)\cap\psi^{-1}\left(\left\{1\right\}\right)} \lim_{n\in\mathbb{N}}\psi\cdot g_n\geq \int_{f^{-1}\left(\left\{0\right\}\right)\cap\psi^{-1}\left(\left\{1\right\}\right)} 1\cdot M\geq M\cdot\lambda\left(f^{-1}\left(\left\{0\right\}\right)\setminus\psi^{-1}\left(\left[0,1\right)\right)\right)\geq M\cdot \frac{\epsilon}{3}. \end{equation}
Therefore $\lambda\left(f^{-1}\left(\left\{0\right\}\right)\right)=0.$
The rest of the proof is simple: $\left|\ln\left(f_{n}\right)\right|\rightarrow \left|\ln\left(f\right)\right|$ almost everywhere and since $\left|\ln\left(f\right)\right| \stackrel{L^{p}}{\rightharpoonup}g$ we have that $g=\left|\ln\left(f\right)\right|.$ (This is an excercise in Folland's Book "Real Analysis." It even follows $L^p$ convergence instead of weak convergence.)
Remark: That excercise is extremely usefull for highly non linear minimization Problems with nasty composition operators. It is not a consequence of dominated convergence. It states that for $L^p$ reflexive, $f_n\rightarrow f$ a.e. and $\left\|f_n\right\|_{L^{p}}$ bounded it holds that $f_n\stackrel{L^p}{\rightarrow}f.$