Hello Im trying to prove that a function h: R^n -> R^m is differentialbe if and only if each of the m components hi: R^n -> R is differentiable. I know that i have to use the coordinate projection function and the chain rule for one implication, but im having lots of trouble starting the problem off. thanks A.P.
If all components are differentiable then my guess is that you can use the triangle inequality to show that the function itself is dif'able. For the only if part, you could assume that one of the components is not differentiable while the function is dif'able and hopefully a contradiction will be obvious.
Ok i used the triangle inequality to show that the function itself is differentiable; my work looks like assume (h1, h2, ...... , hm) is differentiable therefore by the triangle inequality, = ||(h1 .... hm)(x) - (h1.....hm)(Xo) - D(h1.....hm)(Xo)(x-Xo)||/ ||x-Xo|| and since h= (h1,h2, ....., hm) then ||h(x) - h(Xo) - Dh(Xo)(x-Xo)|| / ||x-Xo|| now my proof was that the function is differentiable if and only if each component is differentiable. and i assumed that one component was not differentable where the entire fn was differentiable. To hopefully find that it is a contradiction. my work looks like: assume hj is not differentiable since h= (hi,h2,..hj...,hm) = ||(hi,...hj....hm)(x) - (hi,....hj,....,hm)(Xo) - D(hi,...hj,...,hm)(Xo)(x-Xo)|| / ||x-Xo|| and i went on to say that this is <= || hi(x) - hi(Xo) - Dhi(x-Xo) + ....+ hj(x) - hj(Xo) - Dhj(x-Xo) +.... but i assumed hj was not differentiable therefor this is a contradiction, therefor if h is differentiable then the, sum (i=1 to m) hi must be differentiable is this a good proof showing that all components are differentiable.. or am i doing somthing wrong?