Re: Analysis(Topology) with R^n...

mina_world wrote:
"quasi" <quasi@xxxxxxxx> wrote in message news:jhuom31cmh8j9orkn2bfi9m7qe2f0qpsbm@xxxxxxxxxx
On Fri, 21 Dec 2007 21:19:20 -0500, quasi <quasi@xxxxxxxx> wrote:

On Sat, 22 Dec 2007 10:52:58 +0900, "mina_world"
<mina_world@xxxxxxxxxxx> wrote:

Hello sir~

Let A subset R.
Let f : A -> R^n be given by the equation f(a) = (f_1(a), f_2(a), ... ,
f_n(a)),
where f_m : A -> R for each m.

Then the function f is continuous if and only if each function f_m is
continuous.

-----------------------------------------------------
This is trivial by topology theorem.

If............
Let A subset R.
Let f : A -> R^n be given by the equation f(a) = (f_1(a), f_2(a), ... ,
f_n(a)),
where f_m : A -> R for each m.

Then the function f is differentiable if and only if each function f_m is
differentiable.

is this possible theorem ?
Yes.

Either a routine theorem or else just a definition.

For a function f : A --> R, what is your definition of
I meant:

For a function f : A --> R^n, what is your definition of

"f is differentiable"

?

Yes, you're right.
In fact, I'm unfamiliar to derivatives of vector functions.

If r(t) = <f(t), g(t), h(t)> = f(t) i + g(t) j + h(t) k, where f, g and h are
differentiable functions, then
r'(t) = <f'(t), g'(t), h'(t)> = f'(t) i + g'(t) j + h'(t) k

pf)
r'(t) = lim{d->0} [r(t+d) - r(t)] / d

= lim{d->0} [<f(t+d), g(t+d), h(t+d)> - <f(t), g(t), h(t)>] / d

= lim{d->0} <{f(t+d) - f(t)}/d, {g(t+d) - g(t)}/d, {h(t+d) - h(t)}/d}>

= <lim{d->0}{f(t+d) - f(t)}/d, lim{d->0}{g(t+d) - g(t)}/d,
lim{d->0}{h(t+d) - h(t)}/d}>

= <f'(t), g'(t), h'(t)>

I think all that proves is that if each of the component function si differentiable, then the vector function is differentiable (and the components of the derivative are the derivatives of the components.

To go in the other direction, you have to prove that if the limit of the vector sum exists, then each of the component limits exist, which of course is not very hard. Remember though that it is not generally true that if a limit of a sum exists then the limit of each component of the sum exists.

There is a definition more generally of "differentiable" for functions from R^n -> R^n. For such functions, it is possible for each component function to be differentiable but for the function itself not to be differentiable.



Thank you very much.


.

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