0
$\begingroup$

I'm interested in continuous functions from $\mathbb{R}^n$ to $\mathbb{R}$ that fail to be analytic at a given point (let's say the origin), while still being analytic in a region surrounding it.

Obviously, by definition, I can't expand such a function as a power series around the origin. However, I'm interested in whether there is some other general way in which such functions can be approximated around the origin.

For example, there are several functions that arise in statistical mechanics that can be approximated as

$$ f(x) = \begin{cases} x^\alpha \quad &\text{if $x>0,$} \\ (-x)^\beta \quad &\text{if $x<0$,} \end{cases} $$

for some non-integer $\alpha$, $\beta$, which may or may not be equal. Is this true in general for any function with a non-analyticity at the origin?

If so, I'm interested in what the generalisation looks like to multiple variables. In general, using separate cases as above will break the analyticity at other points than the origin, and I can't seem to come up with anything along the lines of $f(x,y)\approx x^{\alpha_1} y^{\alpha_2}$ that avoids producing complex results for non-integer exponents.

(Note that I'm being vague about the definition of "approximate," because I'm fumbling in the dark a little here. Please interpret it in such a way as to make the question a sensible one, if it is possible to do so.)

$\endgroup$
0
$\begingroup$

I have been interested in this question since you have asked it and now I see that there is little hope for someone else's answer, so I have to try to answer it myself.

Of course, in one dimension, $$ f(x) = \begin{cases} a x^\alpha \quad &\text{if $x>0,$} \\ b (-x)^\beta \quad &\text{if $x<0$,} \end{cases} $$ is another possibility.

Anyway, one could formulate your hypothwsis as: $$ \exists_a \exists_\alpha \lim_{x \rightarrow 0^+} \frac{f(x)}{a x^\alpha} = 1 $$ and $$ \exists_b \exists_\beta \lim_{x \rightarrow 0^-} \frac{f(x)}{b x^\beta} = 1 $$

However, there are other possibilities, such as $f(x) \approx a x^{\alpha_1} |\ln |x||^{\alpha_2}$ (with analogical two limits of ratios — I hope you all see it). You can add $|\ln |\ln |x|||$ and so on. I suspect that there are other possibilities, but I am not sure.

If you want explicitely something in $C^\infty(\mathbb{R}/{0})$, you must of course take into account that $|\ln x|^\alpha$ is not usually continous at 1, so the whole $f(x)$ an a positive half-line can not be the same as the limit form, but something like $|\ln [(\arctan x)/2]|^\alpha$ ($\approx |\ln x - \ln 2|^\alpha \approx |\ln x|^\alpha$ near $x=0$).

In two dimensions, there can be at least functions of the form of $a(\phi) r^{\alpha_1 (\phi)} |\ln r|^{\alpha_2 (\phi)}...$, where $r$ and $\phi$ are polar coordinates, that is $r = \sqrt{x^2 + y^2}$ and $\phi = \operatorname{atan2} (y,x)$ (i.e. $\arctan (y/x)$ or something like this) (and $a(\phi)$ and $\alpha_i (\phi)$ are of course smooth an periodic with period $2 \pi$).

A full example can be $$ f(x) = 2 \sqrt{2} (x^2+y^2)^\frac{1}{4-\frac{0.1 x}{\sqrt{x^2+y^2}}} \left| \ln \left[ \frac{1}{3} \arctan(x^2+y^2) \right] \right|^{2+\frac{x+y}{\sqrt{x^2+y^2}}} \cdot $$ $$ \cdot \left| \ln \left| \ln \left[ \frac{2}{e \pi} \arctan\sqrt{x^2+y^2} \right] \right| \right|^{3+\frac{x+2 y}{\sqrt{x^2+y^2}}} $$

If something is incorrect or unclear, please let me know.

$\endgroup$

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.