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I am reading P.M.H. Wilson's Curved Spaces and would be grateful for some clarification.

Firstly, on page 77,

Of particular importance will be the case when $U$ is an open subset of $\mathbb R^n$ and $f:U\to \mathbb R$ is smooth. For each $P\in U$, we have the linear map $df_P:\mathbb R^n\to\mathbb R$ (an element therefore of the dual space to $\mathbb R^n$). These then yield a smooth map $df:U\to$ Hom$(\mathbb R^n, \mathbb R)$, where Hom$(\mathbb R^n, \mathbb R)$ denotes the dual vector space, consisting of homogeneous linear forms on $\mathbb R^n$, where the dual space may also be identified with $\mathbb R^n$.

I cannot understand what it is saying. I have tried looking up dual space but it's all Greek to me. Also why is this statement true? I understand that if $f:U\to \mathbb R$ is smooth, then we can differentiate it, but why is the resulting $df_P$ necessarily linear and why is $df$ homogeneous? (In my understanding, homogeneous means $f(ax)=a^nf(x)$, for some $n$ right?)

Secondly, on the same page,

$f(x+iy)=u(x,y)+iv(x,y)$ is an analytic function of $z=x+iy$ if and only if the Cauchy-Riemann equations are satisfied.

I remember being told once that there are exceptions, in that all analytic $f$'s satisfy CR but satisfying CR is not sufficient to claim that $f$ is analytic. Is my memory faulty? When does it fail? If I am not mistaken, why can we assume this to be true? Are the example that fail very exotic hence ignorable?

Thank you for your time.

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    @ZhenLin: Thanks for commenting. Hmmm... So they really aren'tall that exotic afterall?2012-02-22

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