# Limiting argument when proving inequality in Sobolev space

I found this limiting argument very common in proving inequalities in Sobolev spaces. Basically, what people do is to observe that test functions (smooth functions with compact support) are dense in $W^{k,p}(\mathbb{R})$, for $1\leq k<\infty, 1\leq p<\infty$, prove the inequality for test functions and then pass to limit.

My concern is: I don't know why the density fact will give us what we want. Could somebody walk me (rigorously) through maybe just one example?

Let's look at this one:

Show the space $W^{1,1}(\mathbb{R})$ embeds continuously into $L^{\infty}(\mathbb{R})$.

It's essentially to show that $||f||_{L^{\infty}(\mathbb{R})}\leq ||f||_{W^{1,1}(\mathbb{R})}$ for any $f\in W^{1,1}(\mathbb{R})$. Proving the inequality for test functions is easy. Then how does that imply the inequality for any $f\in W^{1,1}(\mathbb{R})$?

I suppose, for any $f\in W^{1,1}(\mathbb{R})$, you take a sequence of test functions $f_n$ that converges to $f$ in $W^{1,1}(\mathbb{R})$. We know that $||f_n||_{L^{\infty}(\mathbb{R})}\leq ||f_n||_{W^{1,1}(\mathbb{R})}$. Then take limit. RHS conveges to $||f||_{W^{1,1}(\mathbb{R})}$. But why would LHS converge to $||f||_{L^{\infty}(\mathbb{R})}$?

$$|f_n - f_{n+p}|_{L^\infty} \le |f_n - f_{n+p}|_{W^{1,1}}$$because of the inequality you have shown, hence the sequence $(f_n)$ is a Cauchy sequence in the space $L^\infty$.
The uniqueness of the limit for both norms (maybe because of the convergence in distribution, which is weaker than both convergences, but I am not sure of this point) ensures that $|f_{n+p}|_{L^\infty} \to |f|_{L^\infty}$.
• first thank you for pointing out the Cauchy sequence part. I was stupid to not have seen this. But why does $f_n$ necessarily converge to this same $f$ in $L^{\infty}$? – henryforever14 Mar 29 '14 at 23:14
• So I should interpret this inequality as: For any $f$ in $W^{1,1}$, there is a $\tilde{f}$, s.t. $\tilde{f}=f$ in the distributional sense with $\tilde{f}\in L^{\infty}$ and that inequality holds. Am I right? – henryforever14 Mar 29 '14 at 23:20