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I'm having some problems understanding the following paragraph, which I read in a analysis script (hopefully I haven't made any translation errors):

"A map $f:U \rightarrow Y$, where $U$ is open and $X,Y$ are Banach spaces, is continuous at $x' \in U$ if $$f(x')=\lim_{x\rightarrow x'} f(x)=\lim_{h\rightarrow 0} f(x'+h),$$ where $h=x-x'$. We can decompose $h$ in a "polar" fashion in $h=ts$, where $\left\Vert h \right\Vert \geq 0$ and $s=\frac{1}{\left\Vert h \right\Vert } h$. Then $f(x')=\lim\limits_{h\rightarrow 0} f(x'+h)$ iff $f(x'+ts)\rightarrow f(x')$ for $t \rightarrow 0^+$ uniformly with respect to $\left\Vert s \right \Vert = 1$. No matter from which direction $s$ with $\left\Vert s \right\Vert=1$ we approach $x'$, the value of the function has to converge to $f(x')$ with a to all $s$ common "minimal speed" ".

What I don't understand is this:

1) What does in means to decompose anything in a "polar" fashion ?

2) I thought only sequences of function can converge uniformly...and what does it mean, if something converges uniformly with respect to another thing ?

3) What does the author mean with "common "minimal speed""

Thanks in advance.

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Could you provide the source in any case? – Arturo Magidin May 21 '11 at 20:35
yes, here it is: (it is from a script a lecturer gave us; the stuff which I enclosed in the red box is what I posted here.) – resu May 23 '11 at 14:13
up vote 4 down vote accepted

I guess in (1) it should be $t=\|h\|$, so the analogy would be with the polar decomposition of a complex number $z=ts$ where $t=|z|$ and $s=z/|z|$.

In (2), the uniform convergence probably means that $\sup\limits_{\|s\|=1} \|f(x'+ts)-f(x')\|\to 0$ as $t\to 0_+$.

I suppose the minimal speed thing is meant to be a way of understanding the uniformity condition in (2). One way to keep convergence but violate this uniform convergence would be if you could let $h$ approach $0$ along different paths $\gamma_1,\gamma_2,\dots,$ say with $\|\gamma_i(t)\|=\|\gamma_j(t)\|$ for every $i,j$, but so that the rate of convergence of $f(x'+\gamma_i(t))$ to $f(x')$ gets slower and slower as $i$ increases. So by saying that there should be a lower bound on such rates of convergence (a "common minimal speed", I think) you'd avoid that issue.

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that seems make sense, for the minimal speed... thanks a lot – resu May 23 '11 at 14:15

I'll only deal with (1). Like mac, I'm not sure what (3) means here.

(1) When working in polar coordinates, every point in the plane is written in the form $(r,\theta)$, where $r$ denotes the magnitude (distance to the origin) and $\theta$ denotes the argument (direction). Similarly, when we work with complex numbers, it is often very useful to write a complex number as if it were given by "polar coordinates" rather than "rectangular coordinates". The usual expression $z = a+bi$ with $a,b\in\mathbb{R}$ corresponds to the rectangular coordinates, with $a$ giving the $x$ coordinate and $b$ giving the $y$ coordinate (this is how Hamilton reified the complex numbers); then we can express $z$ instead in "polar coordinates", by writing $z = re^{i\theta}$, where $r$ is a nonnegative real, $\theta$ is a real number, and $$e^{i\theta} = \cos\theta + i\sin\theta$$ gives the "direction." This is often called a "polar decomposition" of the complex number. The number $\theta$ is the argument, and the number $r$ is the magnitude (note that $\lVert z\rVert = r$).

By analogy, in many other circumstances where we express a quantity/object/function/transformation in terms of a "size" and a "direction", we call such an expression a "polar decomposition". This can be done for real numbers very easily: for any real number $h\neq 0$, we can write $h$ as $ts$, where $s$ is the "sign" ($s=1$ if $h\gt 0$, $s=-1$ if $h\lt 0$) and $t$ is the "magnitude" ($t=|h|$). This is the decomposition that is done here.

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great explanation for polar decomposition, thanks. – resu May 23 '11 at 14:14

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