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 $|f_{n+p}|_{L^\infty} \to |f|_{L^\infty}$.

  • $\begingroup$ 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 $L^{\infty}$? $\endgroup$ – henryforever14 Mar 29 '14 at 23:14
  • $\begingroup$ this is true when there is a convergence implied by both metrics. here the weak convergence against test functions (ie, convergence in the sense of distributions) should make the job. $\endgroup$ – mookid Mar 29 '14 at 23:18
  • $\begingroup$ 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? $\endgroup$ – henryforever14 Mar 29 '14 at 23:20
  • $\begingroup$ @henryforever14: yes, this is exactly what I mean. $\endgroup$ – mookid Mar 29 '14 at 23:23
  • $\begingroup$ this makes a hell lot more sense!! Thanks! $\endgroup$ – henryforever14 Mar 29 '14 at 23:25

What I found most convincing is to first observe that both sides of an inequality are continuous with respect to the norm(s) involved. Then this will say that if we have a sequence of functions for each of which a certain inequality holds, then, the same inequality will hold for the limit of the sequence.


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