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Let $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ be two separable Banach spaces. Consider the space of continuously differentiable functions mapping $X$ to $Y$; i.e. $C^1(X, Y)$. Consider the usual $C^1$-topology on this space; i.e. the one constructed through the supremum norm.

Is $C^1(X, Y)$, equipped with this usual $C^1$-topology, Polish? If not, what additional assumptions on $X$ or $Y$ do we need to make it Polish?

Note: To ensure that the supremum norms are bounded, one generally assumes that $X$ is compact in such a problem. Note, however, that I haven't made that assumption from the get-go. This is because, in light of this discussion, I am wondering whether we can somehow avoid making that assumption.

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iheap
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  • To speak of continuously differentiable functions you need more structure than just metric spaces. If you mean instead just continuous functions you can take the following metric which induces uniform convergence: $D(f,g)=\sup\lbrace \min\lbrace d_Y(f(x),g(x)),1\rbrace: x\in X\rbrace$. I do not think that $C(X,Y)$ is always separable. – Jochen Mar 21 '14 at 09:55
  • @Jochen: You are right. I was trying to generalize a specific problem, and going all the way into metric spaces, I went overboard. Let's make $X$ and $Y$ separable Banach spaces. That should fix the differentiability issue, yes? – iheap Mar 22 '14 at 02:36

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No.

First of all, you need to restrict yourself to bounded $C^1$ functions with bounded derivative (which we might denote by $BC^1(X,Y)$, otherwise the $C^1$ "norm" may be infinite.

But consider $X=Y=\mathbb{R}$. I claim $BC^1(\mathbb{R}, \mathbb{R})$ is not separable, for the same reason that $BC(\mathbb{R}, \mathbb{R})$ is not separable. Specifically, for a set $A \subset \mathbb{Z}$, let $f_A \in BC^1(\mathbb{R}, \mathbb{R})$ with $f_A(n) = 1$ for $n \in A$ and $f_A(n) = 0$ for $n \in \mathbb{Z} \setminus A$. (It's clear that we can construct such $f_A$, indeed we can do it with $0 \le f_A \le 1$ and $|f_a'| \le 2$.) Now for $A \ne B$ we have $\|f_A - f_B\|_{C^1} \ge \|f_A - f_B\|_\infty \ge 1$. Therefore by the triangle inequality, the uncountably many open balls $B(f_A, 1/2)$, where $A$ ranges over $2^\mathbb{Z}$, are pairwise disjoint, and therefore any dense subset of $BC^1(\mathbb{R}, \mathbb{R})$ must be uncountable.

Nate Eldredge
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