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I'm having some trouble with the following problems.

Suppose $U \subseteq \mathbb{R}^n$ is open, and suppose ${f}:U \rightarrow \mathbb{R}$ is differentiable at some point $x_0 \in U$.

(1) Show that there exists $\epsilon > 0$ and $M > 0$ such that, for any $h\neq0$ with $|h|<\epsilon$, we have $x_0 + h \in U$ and $|{f}(x_0 + h)-{f}(x_0)|

(2) Show that the following is false: there exist $\epsilon>0$ and $M>0$ such that if $|x-x_0|<\epsilon$ and $|y-x_0|<\epsilon$, then $|{f}(x)-{f}(y)|

Here's my attempt at (1). Clearly there exists $\epsilon$ so that the $\epsilon$-ball about $x_0$ is contained in $U$ since $U$ is open. Set $\gamma:[0,1]\rightarrow U$ defined by $\gamma(t)=t(x_0 + h)+ (1-t)x_0$. Then set $g(t)={f}(\gamma(t))$, and apply the Mean Value Theorem to find a point $t_0 \in (0,1)$ so that $g'(t_0)=g(1)-g(0)={f}(x_0+h)-{f}(x_0)$. However, I don't feel confident about this, because we only have that ${f}$ is differentiable at the point $x_0$, not necessarily on the entire range of $\gamma$.

Any assistance would be greatly appreciated.

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    $(1)$ is false; take $h=0.$ – 2017-02-01
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    Also, why use $\bar f$ instead of $f?$ – 2017-02-01
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    @zhw The notation is not mine. – 2017-02-01
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    Oh, I understand your original comment. (for some reason, I saw $h=0$ and read $h=9$...) I'll amend my post to add the new hypothesis that $h \neq 0$. – 2017-02-01
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    You can always change the notation. It's kind of strange as it is. Suppose I wrote $\tilde f (-\overline {z}) $ for the function? – 2017-02-01
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    Okay, if you so strongly prefer. – 2017-02-01

1 Answers 1

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By differentiability, there is $\epsilon >0$ such that if $0<|h|<\epsilon$, then:

$$|f(x_0+h)-f(x_0) - \nabla f(x_0) \cdot h| < |h|$$

Hence:

$$|f(x_0+h)-f(x_0)| - |\nabla f(x_0)\cdot h| < |h| \\ \therefore |f(x_0+h)-f(x_0)| < |\nabla f(x_0)\cdot h| + |h| \le \|\nabla f(x_0)\| |h| +|h| = (\|\nabla f(x_0)\| +1)|h|$$

Take $M=\|\nabla f(x_0)\| + 1$

For the second part, I suppose what you have to do is e.g. find $f: \Bbb R \to \Bbb R$ which is differentiable at a point (say $0$) but not continuous in any neighborhood of the point. Try:

$$f(x) = \begin{cases} x^2, \text{ if $x\in \Bbb Q$} \\ 0, \text{ otherwise} \end{cases}$$

I haven't yet checked if this works, though.

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    Thanks for your help! I actually thought of a very similar example for (2): $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $f(x)=x$ for $x \in \mathbb{Q}$ and $f(x)=0$ elsewhere. – 2017-02-01