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- Thread starter aurorasky
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Hi aurorasky!

This is indeed true! Take an arbitrary linear map [itex]f:\mathbb{R}^n\rightarrow \rightarrow{R}^m[/itex]. Then it suffices that the components f_{1},...,f_{m} are continuous. And this is true since

[tex]\begin{eqnarray*}

|f_1(x_1,...,x_n)|

& = & |x_1f_i(1,0,...,0)+...+x_nf_i(0,...,0,1))|\\

& \leq & |x_1||f(1,0,...,0)|+...+|x_n||f(0,...,0,1)|\\

& \leq & \|(x_1,...,x_n)\|_\infty(|f(1,0,...,0)|+...+|f(0,...,0,1)|)\\

\end{eqnarray*}[/tex]

which implies continuity. For differentiability, we have to find a linear map u such that

[tex]\lim_{h\rightarrow 0}{\frac{f(x_0+h)-f(x_0)-u(h)}{\|h\|}}=0[/tex]

take u=f and it's easy to see that this is true.

This is indeed true! Take an arbitrary linear map [itex]f:\mathbb{R}^n\rightarrow \rightarrow{R}^m[/itex]. Then it suffices that the components f

[tex]\begin{eqnarray*}

|f_1(x_1,...,x_n)|

& = & |x_1f_i(1,0,...,0)+...+x_nf_i(0,...,0,1))|\\

& \leq & |x_1||f(1,0,...,0)|+...+|x_n||f(0,...,0,1)|\\

& \leq & \|(x_1,...,x_n)\|_\infty(|f(1,0,...,0)|+...+|f(0,...,0,1)|)\\

\end{eqnarray*}[/tex]

which implies continuity. For differentiability, we have to find a linear map u such that

[tex]\lim_{h\rightarrow 0}{\frac{f(x_0+h)-f(x_0)-u(h)}{\|h\|}}=0[/tex]

take u=f and it's easy to see that this is true.

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BTW, could you explain to me how to type latex code?

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Thanks! Also, if a map is differentiable, it is also continuous and so the proof can be made really simple, right?

Yes, but the proof that a differentiable map is continuous uses that linear maps are continuous... (at least the proof I've seen)

BTW, could you explain to me how to type latex code?

Click on the LaTeX symbols to see what I did.

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mathwonk

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then you have to give the differentiable structure, which again is rpesumably the one defined by a linear isomorphism with R^n. Now of coiurse for it to be meaningful, you need to know that every linear isomorphism defines the same differentiable structure, which is equivalent to your original question for R^n, i,.e is every linear map on R^n differentible,

well by definition a map if differentiable if locally it is approximated very well by a linear map, so duhh,... yes a linear map is differentiable.

more precisely a map f is tangent to zero if |f(x)|/|x|-->0 as x-->0. and a map g is differentiable at a if there is a linear map L such that the map g(a+x)-g(a)-L(x) is tangent to zero. Obviously if g is linear and we take L = g, then

g(a+x)-g(a)-L(x) = 0, hence is certainly tangent to zero.

in infinite dimensions some linear maps are not (norm) continuous, and differentiability may be defined as "approximable by a continuous linear map". is this your confusion? i.e., is some nincompoop teaching you in finite dimensional differential calculus before finite dimensional calculus? (that happened to me.) (I was the nincompoop.)

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