Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\mathcal{A}\subset L^p(\mathbb{R})$ is an algebra of functions with the following property:

For every compact $K\subset\mathbb{R}$, $\mathcal{A}$ is dense in $\mathcal{C}(K)$ with respect to the uniform norm $\|\cdot\|_{\infty}$, where $\mathcal{C}(K)$ is the collection of real continuous functions on $K$. The uniform norm I'm referring to is $$\|f\|_{\infty}=\sup_{t\in K}|f(t)|.$$ (See

Can we conclude that $\mathcal{A}$ is dense in $L^p(\mathbb{R})$ (with respect to the $L^p$ norm)?

I became interested in this question while investigating a special case, the $L^2$-density of finite linear combinations of Gaussians:


The question above occurred to me because I can imagine it being useful in cases like this to, say, verify the hypotheses of the Stone-Weierstrass theorem for a given family of functions rather than to explicitly approximate functions in $L^p$.

share|cite|improve this question
I have intuition that in your case it is true that the set is dense, although I have tried attempt at a proof and some problems argue in the opposite direction ; I don't think this is true in the general case. – Patrick Da Silva May 18 '11 at 3:52
It is definitely true in the specific case, that part was homework (last week's). – Nick Strehlke May 18 '11 at 4:08
Yeah, I hope so, but your specific case looks "very nice". – Patrick Da Silva Jun 2 '11 at 4:45

I think you can pull an argument like this : define the sequence of compacts $[-n, n]$ ; since $\mathcal A$ is dense in each of them, for each function $f \in L^p(\mathbb R)$ and for all $n$, there exists a sequence $f_{n,m}(x)$ such that $f_{n,m} \to f$ in $[-n,n]$ with respect to the norm you've chosen as $m \to \infty$.

You could extract a subsequence out of the sequences $f_{n,m}$ by the following argument. Choose a sequence $\varepsilon_k \to 0$. We will find $g_k$ such that $$ \|g_k - f\|_{\infty} \overset{def}{=} \inf \{ K > 0 \, | \, \mu(\{x \in \mathbb R \, | \, |g_k(x) - f(x)| > K\} ) = 0 \} < \varepsilon_k. $$ (I believe this is the detailed definition of your norm? Correct me if I'm wrong.)

with $\mu$ being the Lebesgue measure.

Now $f \in L^p$, so that for $n$ big enough, $|f(x)| < \varepsilon_k/2$ almost everywhere outside $[-n,n]$. Fix this $n$.

This is the point where I am stuck at. Since $f_{n,m} \to f$ in $[-n,n]$, there exists $m$ such that $\|f_{n,m}-f \|_{\infty} < \varepsilon_k$ inside $[-n,n]$ and $|f_{n,m}(x)| < \varepsilon_k/2$ outside$^{*}$ $[-n,n]$ ($f_{n,m} \in L^p$ also, hence goes to $0$ at infinity.)

This guy ($^{*}$) is the problem ; I don't have any control over what happens to $f_{n,m}$ outside $[-n,n]$ and that is what gives me intuition that it might be false in the general case. Although your choice of functions exhibit more properties than the abstract one... but this is now at a state of intuition only. Let me finish just to see where I froze in my argument.

With those properties summed up, outside $[-n,n]$, $|f_{n,m}(x) - f(x)| \le |f_{n,m}(x)| + |f(x)| < \varepsilon_k$ almost everywhere outside $[-n,n]$ and $|f_{n,m}(x) - f(x)| < \varepsilon_k$ inside $[-n,n]$, so that $|f_{n,m}(x)-f| < \varepsilon_k$ almost everywhere over $\mathbb R$, that is, we have found $f_{n,m}$ such that $\|f_{n,m} - f\|_{\infty} < \varepsilon_k$. Just let $g_k = f_{n,m}$.

I hope it helped in some way to at least give you some ideas.

EDIT : I forgot the "$ = 0$" in my definition of the norm. Typo error.

share|cite|improve this answer
I'm going to read this in detail, but the uniform norm I'm referring to is $\|f-g\|_{\infty}=\sup_K|f-g|$. Like this: I apologize for that; I'm going to edit it now to say this explicitly. – Nick Strehlke May 18 '11 at 4:03
Also, to clarify, I meant "Can we conclude that $\mathcal{A}$ is dense in $L^p(\mathbb{R})$ with respect to the $L^p$ norm." I've edited the question to make this clear. – Nick Strehlke May 18 '11 at 4:17
@Patrick Your (*) is about where I am stuck as well. I'm going to edit the question to add some ideas I've had and received so far. Thanks, and let me know if you think of anything else. – Nick Strehlke May 18 '11 at 4:33
You don't need to say that $\mathcal A$ is dense in $\mathcal C(K)$ ; cause it is dense in $\mathcal C(K)$ iff it is dense in $L^p$... just suppose them dense in $L^p$ for less confusion. Also, the supremum norm in $L^p$ is not the same as the supremum norm in $\mathcal C(K)$; you need to use the "essential supremum" instead, which only cares about "the sets that have important points in them" (i.e. if only a singleton is sent "extremely high" by $f$, the essential supremum will remain unchanged. That is the point about the definition of the supremum norm I've stated. – Patrick Da Silva May 18 '11 at 5:38
@Patrick I'm not sure I understand; I'm interested in using properties of $\mathcal{A}$, regarded as a subset of the metric space $(\mathcal{C}(\mathbb{R}),\|\cdot\|_{\infty})$, to prove that it has certain properties when regarded as a subset of $(L^p(\mathbb{R}),\|\cdot\|_{p})$. For example, in the sense that I mean, a function in $\mathcal{C}(K)$ need not even be $p$-integrable over $\mathbb{R}$; $f(x)=x\in\mathcal{C}(K)$ for every $K$, for instance. Please let me know if I'm misstating something here. – Nick Strehlke May 18 '11 at 6:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.