# Proof of equivalence of Sobolev Space and Lipschitz functions

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.

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.

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 $\varphi_i^{h_j}$?
• 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. – Lucio D Mar 17 '14 at 16:10
• 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? – Tomás Mar 17 '14 at 16:19
• $\varphi_{i}^{h_{j}}(x) = \frac{\varphi(x+h_{j}e_{i})-\varphi(x)}{h_{j}} \leq C$ let $g := CM$ then $g \in L^{1}(V)$. – Lucio D Mar 17 '14 at 16:32