Abstract Nonsense

Crushing one theorem at a time

Differentiability Implies Continuity

Point of Post: In this post we prove the obvious fact that if a mapping f:\mathbb{R}^n\to\mathbb{R}^m is differentiable at point it is continuous at that point.

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We would hope that if the total derivative is as similar to the normal derivative as we’d hope that many of the nice qualities pass over. In fact, most of them do. We prove the obvious one here that if something is differentiable (doesn’t ‘blow up’) at a point that it should be continuous (not tear) at that point.

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Differentiability Implies Continuity

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We have no preliminaries, and so we jump to the theorem:

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Theorem: Let f:U\to\mathbb{R}^m where U\subseteq\mathbb{R}^n be differentiable at some point a\in U. Then, f is continuous at a.

Proof: We merely note that since f is differentiable at A we have that for h sufficiently close to \bold{0} that

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\left\|f(a+h)-f(a)-D_f(a)(h)\right\|\leqslant \|h\|

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and so using the reverse triangle inequality we see that

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\left\|f(a+h)-f(a)\right\|\leqslant \|h\|+\|D_f(a)(h)\|\leqslant \left(1+\|D_f(a)\|_\text{op}\right)\|h\|

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where \|\cdot\|_{\text{op}} is the operator norm. Taking the limit as h\to\bold{0} of both sides gives \displaystyle \lim_{h\to\bold{0}}\left(f(a+h)-f(a)\right)=\bold{0} from where the conclusion follows. \blacksquare

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1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.


May 24, 2011 - Posted by | Analysis | , , , , , ,


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