# Does the p-norm converge to the max-norm in some norm

Does the $p$-norm on $\mathbf{R}^n$ converge to the max-norm on $\mathbf{R}^n$ as elements in the space of real valued continuous functions on $\mathbf{R}^n$ endowed with some norm?

More precisely, consider $V = C(\mathbf{R}^n, \mathbf{R})$. Does there exist a norm $\Vert \cdot \Vert$ on $V$ such that the sequence $(\Vert \cdot \Vert_p)_p$ converges to the maximum norm $\Vert \cdot \Vert_\infty$ with respect to $\Vert \cdot \Vert$?

Here's the motivation for this question.

In some sense, I though the max-norm should be the limit of the $p$-norms as $p$ goes to infinity. "Taking an $\infty$-th root of the sum of the infinite powers" in some sense should be the maximum norm. I just thought that this could be made precise.

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It's hard to utilize the $p$-norm on arbitrary real valued functions, which may not be measurable or integrable.. –  Ragib Zaman Oct 19 '11 at 23:12
@Ragib: I don't understand your point. The question seems to be: Consider the functions $f_p: \mathbb{R}^n \to \mathbb{R}$ given by $f_p(x) = \|x\|_p$ as elements of the space $X = C(\mathbb{R}^n, \mathbb{R})$ of continuous functions $\mathbb{R}^n \to \mathbb{R}$. Is there a norm on $X$ such that $f_{p} \to f$ with respect to that norm? I don't know (didn't think about it), but I would like to know what the motivation of this question is... –  t.b. Oct 19 '11 at 23:16
Of course, $\|x\|_{\infty} = \lim_{p \to \infty} \|x\|_p$ for all $x \in \mathbb{R}^n$ (even uniformly on compact subsets of $\mathbb{R}^n$). But why do you want to phrase this in terms of a norm on the space of continuous functions on $\mathbb{R}^n$? –  t.b. Oct 20 '11 at 6:21
@t.b.: Have you ever see a use for that identity (for $L^p$ norms for example)? I'm curious where that might turn out to be useful. –  Jonas Teuwen Oct 20 '11 at 9:05
@JonasTeuwen it can be quite useful in PDE. For example, take the heat equation $u' = \Delta u$, $u(0) = f$. For $2 \leq p < \infty$ simple multiplier methods yield $\|u(t)\|_p \le \|f\|_p$. Taking limits the result holds for $p = \infty$ as well. –  user12014 Nov 24 '11 at 9:26

## 1 Answer

As t.b. noted, we are looking at the functions $f_p:{\mathbb R}^n\to {\mathbb R}$ given by $f_p(x)=\|x\|_p$ and want to know if they converge to $f_\infty$ in some norm on $C({\mathbb R}^n,{\mathbb R})$. It is easy to see that $f_p \to f_\infty$ uniformly on compact sets, and so if you consider the space $C(\Omega,{\mathbb R})$ where $\Omega \subset {\mathbb R}^n$ is bounded and open, then $f_p \to f_\infty$ in, for example, all $L^p$ norms on $C(\Omega,{\mathbb R})$.

For $C({\mathbb R}^n,{\mathbb R})$, the problem is that the the functions $f_p$ do not belong to $C({\mathbb R}^n,{\mathbb R})$, when you endow it with any of the standard norms, such as $L^p$ norms, so we cannot even talk about convergence to $f_\infty$. That being said, you can look at less common norms such as

$\|f\|_X = \sup_{r >0}\left( e^{-r}\sup_{|x|\leq r} |f(x)|\right)$

or weighted $L^P$ norms

$\displaystyle \|f\|_{w,p} = \left( \int_{{\mathbb R}^n} w(x) |f(x)|^p dx\right)^{1/p}$

where $w$ is positive weighting function which decays to zero (exponentially) as $|x|\to \infty$ in order to "cancel out" the growth of $f$. Then the sequence $f_p$ would actually belong to the space $C({\mathbb R}^n,{\mathbb R})$ when endowed with either of these norms. I haven't written this down, but it looks like it would be very easy to show that $\|\cdot\|_p \to \|\cdot\|_\infty$ in both $\|\cdot\|_X$ and $\|\cdot\|_{w,p}$

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The problem with your norms is that these are not norms on all of $C(\mathbb{R}^n,\mathbb{R})$, but rather on some subspaces. Of course, I agree that this is probably the best possible answer to this question if you want to have any kind of natural norm. –  t.b. Oct 25 '11 at 5:31
That's true, I should have mentioned that. But we really only care that all the functions $f_p$ belong to this subspace. –  Jeff Oct 25 '11 at 5:35
Is it even possible to define a norm on $C({\mathbb R}^n,{\mathbb R})$ that applies to all functions in this space? I would be interested to see one if it exists. –  Jeff Oct 25 '11 at 5:37
What you can do is to choose a Hamel basis (algebraic basis) and sum up the absolute values of the coefficients. You can tweak that to a norm in which you have the desired convergence property by taking the subspace $U$ generated by the $p$-norms, endow it with one of those norms you exhibit that work, and endow a vector space complement $V$ with the norm constructed before. Then $\|f\|_U + \|f\|_V$ will be a norm satisfying the requirement of the OP. (that's what I mean by not "natural") –  t.b. Oct 25 '11 at 5:44