# If $f:[a,b] \to \mathbb{R^n}$ is differentiable at $x_0 \in [a,b]$ show that $f$ is continuous at $x_0$.

If $f:[a,b] \to \mathbb{R^n}$ is differentiable at $x_0 \in [a,b]$ show that $f$ is continuous at $x_0$.

I know how to show this for the one dimentional case, but how could I involve uniform convergence to prove this case?

Any help would be appreciated

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If $f$ is differentiable at $x_0$, then for all $\epsilon>0$, there exists a $\delta>0$ such that if $\|x-x_0\|< \delta$, then $\|f(x)-f(x_0)-DF(x_0)(x-x_0)\| \leq \epsilon \|x-x_0\|$. S, choose $\epsilon=1$, then the estimate gives $\|f(x)-f(x_0)\|-\|DF(x_0)(x-x_0)\| \leq \|x-x_0\|$, which can be written as $\|f(x)-f(x_0)\| \leq (1+\|Df(x_0)\|) \|x-x_0\|$. Setting $L = 1+\|Df(x_0)\|$ shows that $\|f(x)-f(x_0)\| \leq L \|x-x_0\|$ locally.

I don't understand what you meant by 'uniform convergence' in your question.

It is possible that the function is only continuous (and differentiable) at one point. For example, take $f(x) = x^2 1_\mathbb{Q}(x)$. Then $f$ is differentiable (and hence continuous) at $x=0$, but is discontinuous (and hence not differentiable) at all $x \neq 0$.

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This is the standard proof for functions from $\mathbb{R}^m$ to $\mathbb{R}^n$: consider what happens to $$f(x+h)-f(x)=|h|\frac{f(x+h)-f(x)-f'(x)h}{|h|}+f'(x)h$$ as $h \to 0$.

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Isn't it the same as my one? –  Dominic Michaelis Feb 15 '13 at 5:53

So far the other answers address the question by appealing to differentiability in the more general context where the domain is $\mathbb R^m$. While this certainly will solve the problem posed, it is a bit of an overkill. The case of a function $f:[a,b]\to \mathbb R^n$ is considerably simpler than when the domain is of higher dimension. In fact, the case of $f:[a,b]\to \mathbb R^n$ can be immediately reduced to the case of single variable real valued functions, as follows.

A function $f:[a,b]\to \mathbb R^n$ is essentially the same as $n$ functions $f_1,\cdots ,f_n:[a,b]\to \mathbb R^n$, the component functions, given by the equality $f(x)=(f_1(x),\cdots ,f_n(x))$. Now, anything you want to know about the function $f$ can be answered by considering its component functions. The following two exercises are good examples of that:

1) $f$ is continuos at $x_0$ if, and only if, each of its component functions $f_i$ is continuous at $x_0$.

2) $f$ is differentiable at $x_0$ if, and only if, each of its component functions $f_i$ is continuous at $x_0$.

The proofs follow quite automatically by carefully following the definitions. Now, since for ordinary single variable real valued functions, differentiability at a point implies continuity at that point, the desired result follows.

Remark: The case where the domain is one dimensional and the codomain is of higher dimension is considerably simpler than when the domain is of higher dimensions and the codomain is of dimension one. For the case at hand, function $f:[a,b]\to \mathbb R^n$ can be thought of as paths in $\mathbb R^n$, and the derivative is then the vector of instantaneous change. The components of the derivative vector are precisely the derivatives of the component functions.

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Take the definition of differentiable: $$f(x+h)=f(x)+A\cdot h+r(h)$$ with $$\lim_{\|h\|\rightarrow 0} \frac{r(h)}{\|h\|}=0$$ Now take a sequence which converges to $f(x)$.

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