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There is theorem that says if a function is differentiable at a point in an open set, then it is continuous at the point. This can be proven by doing :

$$\lim_{x \rightarrow a} f(x)-f(a) = \lim_{x \rightarrow a} (x-a) \left( \frac{f(x)-f(a)}{x-a}\right) \Rightarrow \lim_{x \rightarrow a} f(x) = f(a).$$ The same proof is true in higher dimensions?

So if we let $S \subset \mathbb{R}^n$ and $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ be defined on $S$. Is the proof is true if we just replace $(x-a)$ with $\|x-a\|$??

--Edit

The definition of differentiability that my text book using is the following :

A mapping $f$ from an open set $S \subset \mathbb{R}^n$ into $\mathbb{R}^m$ is said to be differentiable at $a \in S$ if there is an $m \times n$ matrix such that $\lim_{h \rightarrow 0} \frac{\|f(a+h) - f(a) - L h\|}{\|h\|} = 0$ where $L$ is called the Frechet deribative.

eChung00
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  • check https://en.wikipedia.org/wiki/Fréchet_derivative#Definition BY definition $$ f(x + h) = f(x) + Ah +o(h)$$ for $h\to0$ therefore also $$ f(x + h) = f(x) +o(1)$$ that is $f$ is continuous at $x$ – Pietro Majer Oct 15 '14 at 00:25
  • It would be easier to give an answer to this question which is useful to you (eChungOO), if you would provide the exact definition of differentiability that you are using. – PhoemueX Oct 15 '14 at 07:24
  • @PhoemueX Hi, I added the definition that my text book is using.. thanks – eChung00 Oct 15 '14 at 14:33

1 Answers1

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The proof is very similar to the one in the one-dimensional setting, but not completely identical.

Observe that the triangle inequality yields

\begin{eqnarray*} \left\Vert f\left(a+h\right)-f\left(a\right)\right\Vert & = & \left\Vert h\right\Vert \cdot\frac{\left\Vert f\left(a+h\right)-f\left(a\right)\right\Vert }{\left\Vert h\right\Vert }\\ & \leq & \left\Vert h\right\Vert \cdot\frac{\left\Vert f\left(a+h\right)-f\left(a\right)-Lh\right\Vert }{\left\Vert h\right\Vert }+\left\Vert Lh\right\Vert . \end{eqnarray*}

I will let you take it from here.

PhoemueX
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