# Differentiability implies continuity proof

Statement:

If $$f: \mathbb{R}^n\rightarrow \mathbb{R}^m$$ is differentiable at $$\underline{x}$$, it's continuous at $$\underline{x}$$.

My proof:

Since $$f$$ is differentiable at $$\underline{x}$$, all directional derivatives exist.

Note also that for $$\underline{h}=(h_1,\cdots, h_n),$$ $$||f(\underline{x}+\underline{h})-f(\underline{x})|| = ||\sum\limits_{i=1}^n f(\underline{x}+\underline{v_i})-f(\underline{x}+\underline{v_{i-1}})||$$ where $$v_i = (h_1, h_2, \cdots, h_i, 0, \cdots, 0)$$ when $$i\neq 0$$ and $$(0,\cdots,0)$$ when $$i=0$$.

Thus, on an arbitrary direction $$i$$, since the derivative exists, the function reduced to that direction is continuous

Therefore $$\forall \epsilon>0, \exists \delta_i>0$$ such that $$\forall ||(\underline{x}+\underline{v_i})-(\underline{x}+\underline{v_i})||=||(0,\cdots, 0, h_i, 0, \cdots,0)||=|h_i|<\delta_i$$, we have: $$||f(\underline{x}+\underline{v_i})-f(\underline{x}+\underline{v_{i-1}})||<\frac{\epsilon}{n}$$

And finally, let $$\delta = \min\{\delta_i\}$$ and $$\forall \underline{h}\in B_\delta(\underline{x})$$, $$||f(\underline{x}+\underline{h})-f(\underline{x})|| = ||\sum\limits_{i=1}^n f(\underline{x}+\underline{v_i})-f(\underline{x}+\underline{v_{i-1}})||<\frac{\epsilon}{n}\cdot n=\epsilon$$

Question:

I'm not sure if my proof is correct, it is unnecessarily more complicated than the textbook proof but I would be grateful if someone could tell me if this was a right way, and if not, what are the problems about it.

• You may want to try to mimic the 1D proof. – Calvin Khor Feb 23 at 18:06
• Recall that, by sheer definition of differentiability at $x,$ there exists a linear function $u$ such that $f(x + h) = f(x) + u(h) + e(h),$ with $\|e(h)\|/\|h\| \to 0$ as $\|h\| \to 0$ (with $h \neq 0$). Then, make $h \to 0$ and you'll get right away $f(x + h) \to f(x).$ – Will M. Feb 23 at 21:45

No, it is not correct. The restriction of $$f$$ to every straight line being continuous at $$(0,0,\ldots,0)$$ does not imply that $$f$$ is continuous there. Consider, for instance,$$\begin{array}{rccc}f\colon&\mathbb{R}^2&\longrightarrow&\mathbb R\\&(x,y)&\mapsto&\begin{cases}1&\text{ if }y=x^2\text{ and }(x,y)\neq(0,0)\\0&\text{ otherwise.}\end{cases}\end{array}$$