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A function $f: \mathbb{R^n} \to \mathbb{R^m}$ is differentiable at $a$ if there exists a linear map $ \lambda: \mathbb{R^n} \to \mathbb{R^m}$ such that

$$\lim_{h \to 0} \frac{\|f(a+h) - f(a) - \lambda(h)\|}{\|h\|} = 0$$

So clearly, if $f$ is differentiable at $a$ then $\lim_{h \to 0} f(a+h) - f(a) - \lambda(h) = 0$, but where do you proceed from here?

WacDonald's
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3 Answers3

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More explicitly:

The limit shows that for any $\epsilon>0$, there exists a $\delta>0$ so that if $\|h\| < \delta$, then $\|f(a+h) - f(a) - \lambda(h)\| \leq \epsilon \|h\|$. $\lambda$ is continuous, hence bounded, so we have $\|\lambda(h)\| \leq K \|h\|$, for some $K$.

Then we have $\|f(a+h) - f(a)\| \leq \|f(a+h) - f(a) - \lambda(h)\| + \|\lambda(h)\|\leq (\epsilon + K ) \|h\|$.

Now let $\eta >0$, then if $\|h\| < \min(\delta, \frac{\eta}{\epsilon+K})$, we have $\|f(a+h) - f(a)\| < \eta$, which shows that $f$ is continuous at $a$.

copper.hat
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Note that $\lambda$ is linear, so $\lim\limits_{h\to 0} \lambda(h)=0$. Thus $\lim\limits_{h\to 0} f(a+h)-f(a)=0$, which is equivalent to $\lim\limits_{x\to a}f(x)=f(a)$, one of the definitions of continuity (of course all the definitions are equivalent).

Alex Becker
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  • Why is this step justified? – WacDonald's Jul 19 '12 at 02:07
  • @WacDonald's Which one? – Alex Becker Jul 19 '12 at 02:09
  • From what I wrote down in the last line, how can you conclude $\lim_{h \to 0} f(a+h) - f(a) = 0$? (i.e. why are you allowed to distribute the limit)? – WacDonald's Jul 19 '12 at 02:11
  • @WacDonald's Because $\lim\limits_{h\to 0}\lambda(h)=0$. Thus $$\lim\limits_{h\to 0}(f(a+h)-f(a))=\lim\limits_{h\to 0}(f(a+h)-f(a))-\lim\limits_{h\to 0}\lambda(h)=\lim\limits_{h\to 0}(f(a+h)-f(a)-\lambda(h))$$ – Alex Becker Jul 19 '12 at 02:20
  • So, $\lim_{h \to 0} f(a + h) - \lambda(h) = f(a) = \lim_{h \to 0} f(a+h) - \lim_{h \to 0} \lambda(h) $? – WacDonald's Jul 19 '12 at 02:28
  • What's bothering is that just because $\lim_{x \to a} f(x) + g(x) $ exists doesn't mean $\lim_{x \to a} f(x) + g(x) $ = $\lim_{x \to a} f(x) + \lim_{x \to a} g(x)$ – WacDonald's Jul 19 '12 at 02:29
  • @WacDonald's The problem is you're not keeping track of the directions of the implications. If $\lim\limits_{x\to a}F(x)$ and $\lim\limits_{x\to a}g(x)$ exist, then $\lim\limits_{x\to a}(f(x)+g(x))$ exists and is equal to their sum. Hence the derivation I gave in my comments (noting that $\lim\limits_{x\to a} -f(x)=-\lim\limits_{x\to a} f(x)$). – Alex Becker Jul 19 '12 at 02:35
  • Just one question. Does differentiability require $\lim_{h \to 0} f(a + h) - f(a)$ to exist? – WacDonald's Jul 19 '12 at 02:41
  • @WacDonald's You're right, I made an error in my explanation. I meant to write $$0=\lim\limits_{h\to 0} (f(a+h)-f(a)-\lambda(h))=\lim\limits_{h\to 0} (f(a+h)-f(a)-\lambda(h))+\lim\limits_{h\to 0} \lambda(h)=\lim\limits_{h\to 0} (f(a+h)-f(a))$$ – Alex Becker Jul 19 '12 at 02:45
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Since $\lambda(h)$ is a linear mapping from $\mathbb{R}^n$ to $\mathbb{R}^m$, it is continuous and $\lambda(0)=0$. Hence $\displaystyle \lim_{h\rightarrow 0}\lambda(h)=\lambda(0)=0$.

blindman
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