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.
Thanks in advance!
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,$$\begin{array}{rccc}f\colon&\mathbb{R}^2&\longrightarrow&\mathbb R\\&(x,y)&\mapsto&\begin{cases}1&\text{ if }y=x^2\text{ and }(x,y)\neq(0,0)\\0&\text{ otherwise.}\end{cases}\end{array}$$