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A local analysis textbook I have used has the following exercise:

Let $X$ be a finite-dimensional, $Y$ a separable Banach-space, $f\colon X\rightarrowtail Y$ any function. Show that $f'$ is Borel.

I have no idea how to approach this. I looked at the proof of Rademacher, but even that only gives Lebesgue-measurability as far as I can see, and this has no assumptions on the function at all. Any help would be appreciated.

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Out of curiosity (I'm not sure if this statement is true or not): Have you actual need for this statement or is it out of interest that you ask? –  t.b. May 11 '11 at 23:30
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Can you do the case $X = Y = \mathbb R$ ? –  GEdgar May 11 '11 at 23:59
    
I do not need it for an application now, it is more out of interest. –  scineram May 12 '11 at 8:48
    
Even if $X=Y=\mathbb{R}$ I cannot approach this. I do not know what possible method could apply to functions like $id^2_\mathbb{R}\cdot\chi_A$ for any $A$. Should I bring this mathoverflow? –  scineram May 12 '11 at 8:54
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Define $f_n(x):=\frac{f(x + 1/n) - f(x)}{1/n}$... then... –  Jonas Teuwen May 12 '11 at 9:22

1 Answer 1

It seems my professor whose book I originally met this problem in published the proof himself, generalizing the previous result of Federer for continuous functions. I was really surprised how much simpler it was than I expected.

In fact, I noticed the proof only requires that $L(X;Y)$ is separable, hence so are $X'$ and $Y$, since they embed into it. Given these, either of them being finite-dimensional is sufficient. It also raises the question whether $L(X;Y)$ can be separable when both spaces are infinite-dimensional. I read somewhere that this is an open problem when $X=Y$.

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