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The attachment is a proof from Evans book "Measure Theory and Fine Properties of Functions" pg 132 Theorem 5. The statement of the theorem is:

Let $f:U \rightarrow \mathbb{R}$. Then $f$ is locally Lipschitz in $U$ if and only if $f \in W^{1,\infty}_{loc}(U)$. My two questions are only regarding the first direction($\Rightarrow$) of the theorem.

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  1. Can we get the sequence $g_{i}^{h_{j}} \rightharpoonup g_{i}$ weakly in $L^{p}_{loc}(U)$, where $g_{i} \in L_{loc}^{\infty}(U)$ by noting: Since $(g_{i}^{h_{j}})_{j}$ is bounded in $L^{\infty}_{loc}(U)$ we can find a weak* convergent subsequence such that $g_{i}^{h_{j}} \rightharpoonup g_{i}$ in $L^{\infty}_{loc}(U)$ and then use Theroem 3 in section 1.9 (which states: since $g_{i}^{h_{j}}$ is bounded in $L^{\infty}_{loc}(U)$ there exists a subsequence which converges weakly in $L^{p}_{loc}(U)$) to get a further subsequence (which I will again call $g_{i}^{h_{j}}$) such that $g_{i}^{h_{j}} \rightharpoonup v$ weakly in $L^{p}_{loc}(U)$, by uniqueness of limits we get $v = g_{i}$. Does this reasoning work? Is there a simpler way to show this?

  2. To get the result that $\lim\limits_{j \rightarrow \infty}\int_{U}f(x)\frac{\varphi(x+h_{j}e_{i})-\varphi(x)}{h_{j}}dx = \int_{U}f(x)\frac{\partial \varphi}{\partial x_{i}}dx$ at the end, can I use the Lebesgue Dominated Convergence Theorem where I would dominate $f(x)\frac{\varphi(x+h_{j}e_{i})-\varphi(x)}{h_{j}}$. Since $\varphi \in C_{c}^{1}(V)$ is continuous on a compact set it is bounded, so $\frac{\varphi(x+h_{j}e_{i})-\varphi(x)}{h_{j}} \leq C $ for some constant $C > 0$. But since $V$ is bounded we get $\int_{V}Cdx \leq C\int_{V}dx = C|V| < +\infty$. Then $g := MC$ becomes the summable function which dominates , where $M := \Vert f \Vert_{L^{\infty}_{loc}(U)}$. Does this reasoning work? Is there a simpler way to show this?

Please let me know where I went wrong first before showing the correct reasoning.

Thanks for any assistance. Let me know if anything is unclear.

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  1. It seem to me that the argument is fine. The key idea here is that $L^p(U)$ is continuously embedded in $L^q(U)$ for $p\ge q$ ($U$ bounded). Note also that we only need, for example, that $g_i^{h_i}$ has a subsequence wich weakly converge in $L^2(U)$.

  2. I do agree with this argument too, however, you do not finished it. What is the function in $L^1(U)$ which dominates $\varphi_i^{h_j}$?

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  • $\begingroup$ Oh yes I see, will correct. Since $\varphi$ is continuous on a compact set it is bounded, so $\varphi \leq C $ for some constant $C > 0$. But since $V$ is bounded we get $\int_{V}Cdx \leq C\int_{V}dx = C|V| < +\infty$. Then $g := MC$ becomes the summable function which dominates. $\endgroup$ – Lucio D Mar 17 '14 at 16:10
  • $\begingroup$ I can't understand your argument. You have to find a function $g\in L^1(U)$ such that $$\varphi_i^{h_j}(x)\le g(x),\ a.e.\ x\in U,\ \forall j$$ Where did you acc?omplished it? $\endgroup$ – Tomás Mar 17 '14 at 16:19
  • $\begingroup$ $\varphi_{i}^{h_{j}}(x) = \frac{\varphi(x+h_{j}e_{i})-\varphi(x)}{h_{j}} \leq C $ let $g := CM$ then $g \in L^{1}(V)$. $\endgroup$ – Lucio D Mar 17 '14 at 16:32
  • $\begingroup$ Ok @LucioD, now I understood. $\endgroup$ – Tomás Mar 17 '14 at 16:51

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