A Banach Space is a vector space which is normed and is complete with respect to its norm (every cauchy sequence converges). Examples of Banach spaces include $\mathbb{R}^n$ and the space of bounded continuous functions with the norm $\Vert f(x) \Vert = \sup \lbrace \vert f(x) \vert \rbrace$.

The set of all Bounded Linear Operators $\Omega$ on the space of all $C^\infty$ functions over $\mathbb{R}$ forms a Banach space:

  1. For two operators $A, B \in \Omega$ define $(A+B)(f) = Af + Bf$, and for a constant $c \in \mathbb{R}$ define $(cA)(f) = c Af$. This makes $\Omega$ into a vector space.

  2. Define a norm by $ \Vert A \Vert_\Omega = \sup \lbrace \Vert Af \Vert : f \in C^\infty(\mathbb{R}), \Vert f \Vert \leq 1 \rbrace $, where $\Vert f \Vert$ for $f \in C^\infty(\mathbb{R})$ is some norm on that space.

Now my question is: Does the norm chosen for $C^\infty(\mathbb{R})$ affect the structure of $\Omega$ (i.e. its topology or whatever), and is there a natural choice for such a norm in this situation?

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    $\begingroup$ It's not clear to me how you would define a norm on $C^\infty(\mathbb{R})$ at all. $\endgroup$ Aug 15, 2010 at 3:29
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    $\begingroup$ The sup-norm and $L_2$ norm are both norms that work on $C^\infty(\mathbb{R})$. $\endgroup$ Aug 15, 2010 at 4:10
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    $\begingroup$ @Eric Haengel: Why is this tagged under complex analysis? $\endgroup$
    – anonymous
    Aug 15, 2010 at 6:55
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    $\begingroup$ Neither the "sup-norm" nor the "$L_2$" norm is a norm on $C^\infty(\mathbb{R})$. Of course the norm chosen on an infinite-dimensional vector space determines which operators on it are bounded. There are various topologies on $C^\infty(\mathbb{R})$ studied in the theory of distributions en.wikipedia.org/wiki/Distribution_%28mathematics%29 but they are not derived from norms. $\endgroup$ Aug 15, 2010 at 7:40
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    $\begingroup$ Greg, no. The "sup-norm" is not a norm on $C^\infty(\mathbb{R})$ (as I've already said). $\endgroup$ Aug 15, 2010 at 10:03

2 Answers 2


I'm not aware of any natural choices for norms on $C^\infty(\mathbb{R})$ at all. (As mentioned in comments, things like the sup or $L^p$ norms are infinite for some $C^\infty(\mathbb{R})$ functions.)

There are unnatural choices, contrary to Owen Sizemore's assertion. For example, it is known that all infinite-dimensional separable Fréchet spaces (such as $C^\infty(\mathbb{R})$ in its usual topology) have Hamel dimension $2^{\aleph_0}$, hence are isomorphic as vector spaces. (I believe this is due to Mazur, but couldn't find the exact reference.) So if $(X, ||\cdot||_X)$ is your favorite infinite-dimensional separable normed space, there is a linear isomorphism $T : C^\infty(\mathbb{R}) \to X$ (of course it is in general horribly discontinuous). For $f \in C^\infty(\mathbb{R})$, set $||f|| := ||Tf||_X$. Voilà, a norm on $C^\infty(\mathbb{R})$.

Of course, under this norm $C^\infty(\mathbb{R})$ is isometrically isomorphic to $X$, so for all intents and purposes you are really working with $X$. You can even make $C^\infty(\mathbb{R})$ into a Banach space or even a Hilbert space by choosing $X$ accordingly. And the space of bounded operators on $(C^\infty(\mathbb{R}), ||\cdot||)$ is isometrically isomorphic to the space of those on $X$. Thus for any infinite-dimensional separable normed space $X$ you can realize the space of bounded operators $L(X)$ in this way, so the choice of norm on $C^\infty(\mathbb{R})$ certainly does affect the structure of $L(C^\infty(\mathbb{R}))$.

This illustrates in some sense that the norm contains almost all of the structure of a normed space; if you drop it, you are left with just the vector space structure, which determines almost nothing. "Choosing a norm" on a space is not really a sensible thing to do, since in effect it is really just choosing a normed space which could be almost totally unrelated to the original space.

Also, it should be pointed out that the space of bounded linear operators on an incomplete normed space is not a Banach space. Proof: Let $X$ be an incomplete normed space, let $\lbrace x_n\rbrace$ be a Cauchy sequence in $X$ with no limit, and let $f \in X^*$ be a nonzero bounded linear functional on $X$. Define operators $T_n$ by $T_n x = f(x) x_n$. Then $\lbrace T_n \rbrace$ is Cauchy in operator norm, but does not converge.

The correct theorem is that for normed spaces $X,Y$, the space $L(X,Y)$ of bounded linear operators from $X$ to $Y$ is a Banach space iff $Y$ is a Banach space.


First off there is no norm that can be put on $C^\infty(\mathbb{R})$, or even $C^\infty([0,1])$, that will make it into a Banach Space. I mention the last one because there are lots of obvious norms, sup norm, $L^p$ norms etc. But it will not be complete with respect to any norm.

It's fairly simple to see this, for the $L^2$ norm. Differentiation would then be an operator defined on the whole Banach space that has an adjoint. In fact, it is self-adjoint (this is by integration by parts).

Now it is a theorem that any operator on a Banach space with an adjoint is automatically bounded (this is a consequence of Total Boundedness Principle). But of course differentiation is unbounded.

As for your main question:

If you put some norm on $C^\infty(\mathbb{R})$ then yes it would greatly affect the set of bounded operators (even as a set, not just as a banach space). For example putting different $L^p$-norms would give you different duals spaces. While if you say restricted to bounded smooth functions and used the sup-norm, the dual would be measures on $\mathbb{R}$.

I guess the take-away message is that the definition of bounded operators involves the norm on the domain. So if you change that norm you can greatly change the bounded operators.

By the way most of this stuff can be found in pretty much any Functional Analysis textbook.

I like Analysis NOW by Gert Pederson.

  • $\begingroup$ +1 for Analysis NOW, but -1 for "there is no norm that can be put on $C^\infty(\mathbb R)$, that will make it into a Banach Space". So, $\pm 0$. $\endgroup$
    – Rasmus
    Aug 15, 2010 at 14:44
  • $\begingroup$ Ok, yeah so I meant there is no natural norm. Since $C^\infty(\mathbb{R})$ is not countable dimension then it is isomorphic as a vector space to any other separable Banach space, and so you can put a norm on it through the iso. Maybe a good way to say it, is there is no explicit norm. (i.e. without using choice to define). $\endgroup$ Aug 16, 2010 at 16:19
  • $\begingroup$ Do you know a reference, where I can find a proof that, there is no norm in $C^\infty([0,1])$ which makes it a Banach space? $\endgroup$
    – Tomás
    Sep 3, 2014 at 0:18
  • $\begingroup$ @Tomás: The way I usually know involves using the uniform boundedness principal. Since differentiation is self-adjoint, it $C^\infty$ was a Banach space that would mean that differentiation is bounded, which it can't be (no matter the norm as $x^n$) shows. Any intro functional analysis textbook will have the uniform boundedness principle. Though I haven't seen any that explicitly shows that $C^\infty$ is a Frechet space that isn't a Banach space. $\endgroup$ Sep 3, 2014 at 3:53
  • $\begingroup$ Dear @OwenSizemore, I would like to refere you to this post. $\endgroup$
    – Tomás
    Sep 5, 2014 at 14:34

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