0
$\begingroup$

I am reading some paper and I encountered this statement:

... the coefficients $a_{p,\beta}(t,x)$ [are] of class $C^m$ in $t$, valued in the space of analytic functions of $x$, in a neighborhood of the origin...

Just to clarify, $t$ is real 1-dimensional and $x$ is real $n$-dimensional. So the domain of $a_{p,\beta}(t,x)$ is some neighborhood of the origin in $\mathbb{R} \times \mathbb{R}^n$.

I want to be sure of what this statement means. First, $a_{p,\beta}$ being of class $C^m$ in t means that if I fix $x$, then I get an $m$-times continuously differentiable function in $t$. Is this right?

Now, what does it mean for $a_{p,\beta}$ to be valued in the space of analytic functions of $x$? Does this mean that if I fix $t$ this time, I get a function analytic on some neighborhood of the origin in $\mathbb{R}^n$?

$\endgroup$
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – user1337 Aug 7 '13 at 22:39
1
$\begingroup$

So, there is some space $X$ of analytic functions in some (presumably fixed) neighborhood of the origin. An analytic function on an open subset of $\mathbb R^n$ is a function that is locally representable by a power series, or, equivalently, one that extends to an analytic function on an open subset of $\mathbb C^n$.

For every fixed real $t$ near the origin, we get a function $F_t(x) = a_{p,\beta}(t,x)$ which is an element of $X$. This means we have a map from an interval into $X$. Being $C^m$ for such a map means having $m$ continuous derivatives, just as for real-valued functions. The definition of derivative $\lim_{h\to 0} \frac{1}{h}(F_{t+h}-F_t)$ makes sense for maps into any topological vector space: all we need is the vector space structure to form $\frac{1}{h}(F_{t+h}-F_t)$, and a topology to define what a limit is.

Generally speaking, the above is stronger than being $C^m$ in $t$ for every fixed $x$. Being differentiable in $t$ for every fixed $x$ means we have pointwise limits $\lim_{h\to 0} \frac{1}{h}(F_{t+h}(x)-F_t(x))$. But if we think of $F$ as a map into function space $X$, the limit has to exist in the topology of that space, which may be stronger than the topology of pointwise convergence (e.g., topology induced by the uniform norm).

Also, a remark: consider giving references to paper(s) you wish explained. Sometimes the part quoted in the question is not enough to answer unambiguously, while some other part of the paper may have helpful context. For example, if I knew what is happening to $a_{p,\beta}$ later in the paper, I probably would have given a better answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.