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Prove $C^\infty(\Bbb R^n)$ is a Banach Space when equipped with topology of uniform convergence. $C^\infty(\Bbb R^n)$ is space of all continuous functions that converge to $0$ at $\infty$. And, the topology of uniform convergence is defined by the norm $f \mapsto \sup_{x\in\Bbb R^n}|f(x)|$.

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    $\begingroup$ Do you have a question? $\endgroup$ Commented Mar 10, 2013 at 20:10
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    $\begingroup$ Also, it doesn't make sense to ask whether a vector space equipped with a certain topology is a Banach space. To be a Banach space requires being equipped with a norm. $\endgroup$ Commented Mar 10, 2013 at 20:18
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    $\begingroup$ I'd be very curious to see what textbook this could have been copied from. Ordinarily, $C$ in such contexts means continuous functions, $C^\infty$ means infinitely differentiable functions, convergence to zero at infinity is denoted with a subscript $0$ on the $C$. While I typed this, Ty corrected another problem by inserting absolute value signs so that the norm is non-negative. But without a continuity assumption, the first counterexample in Chris Eagle's answer survives. $\endgroup$ Commented Mar 10, 2013 at 20:27
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    $\begingroup$ Note that in many books $C^\infty_0$ does not denote the smooth functions that vanish at $\infty$. It often denotes the smooth functions with compact support! This is strange somehow because $C_0$ means only vanishing at infinity :) $\endgroup$ Commented Mar 10, 2013 at 21:05
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    $\begingroup$ Another addition: It is possible to approximate a mere continuous function uniformly by smooth functions. However the limit is not smooth then. Think of a hat function for example. So the statement is even false for $C^1$ functions since your norm does not control the derivatives $\endgroup$ Commented Mar 10, 2013 at 21:08

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The statement is false. Norms have to be finite, but if we let $n=1$ and take $$f(x)=\begin{cases} 1/x & x \neq 0 \\ 0 & x=0\end{cases}$$ then $f$ is in your space but has infinite norm.

Norms also have to be positive, but if we take $f(x)=-1$ then the norm is $-1$.

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    $\begingroup$ I fixed my two problems I left out continuous functions and forgot to put |f(x)|. Thanks $\endgroup$ Commented Mar 10, 2013 at 20:27
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It's not difficult to show $\|f\|=\sup_{x\in \mathbb{R}^n}|f(x)|$ is a norm. We need to show the space is complete.

Suppose $\{f_n\}_{n=1}^\infty$ is a Cauchy sequence in the space $i.e.\ \|f_n-f_m\|\to0$ as $m,n\to\infty$. For any fixed $x\in \mathbb{R}^n,\ \{f_n(x)\}$ is Cauchy in $\mathbb{R}$(I hope I didn't misunderstand the map $f: \mathbb{R}^n\to\mathbb{R}$).

Define $f(x):=\lim_{n\to\infty}f(x)$, this is a uniform convergence $i.e. \|f_n-f\|\to 0$ as $n\to\infty$. Hence, $f$ is continuous and moreover,

$$ |f(x)|\le |f_n(x)-f(x)|+|f_n(x)|\le \|f_n-f\|+|f_n(x)| $$ $\forall \epsilon>0,\ \exists\ M>0\ s.t.\ \|f_n-f\|\le\epsilon$ when $n>M$. For a fixed $n>M,\ \exists\ N>0,\ s.t.|f_n(x)|<\epsilon$ if $|x|>N$. Therefore, there is an $N$ such that for any $|x|>N$ $$ |f(x)|\le \|f_n-f\|+|f_n(x)|<2\epsilon\ \Rightarrow\ f\ converges\ to\ 0\ at\ \infty $$ So $f$ is an element in your space.

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    $\begingroup$ The uniform limit of an arbitrary smooth function has only to be continuous! With (only) uniform convergence one leaves the space of $C^\infty$-functions as I mentioned above. For example take a just continuous function with compact support, like a "symmetric hat" and convolute it with a standard dirac approximation. The convolution is then $C^\infty$ with compact support thus in the desired space but converges uniformly to a mere (Lipschitz)-continuous function $\endgroup$ Commented Mar 11, 2013 at 6:44
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    $\begingroup$ @Quickbeam2k1 Cann't you see that "$C^{\infty}$" discussed here is a space of continuous functions, rather than smooth functions which is generally understood in analysis? $\endgroup$
    – Coiacy
    Commented Mar 11, 2013 at 7:21
  • $\begingroup$ Oh I'm to sorry. Could you edit your answer slightly so I can upvote it. $\endgroup$ Commented Mar 11, 2013 at 7:37
  • $\begingroup$ @Quickbeam2k1 That's OK, no need to upvote it. Actually I was also confused about the "$C^{\infty}$"at first sight. $\endgroup$
    – Coiacy
    Commented Mar 11, 2013 at 7:40
  • $\begingroup$ but I downvoted it :) maybe you could just add something like often the space is called $C_0(\mathbb{R}^n$) or something like that $\endgroup$ Commented Mar 11, 2013 at 7:42

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