February | 2021 | Tao Analysis Solutions

Exercise statement

Suppose that $X$ is a subset of $\mathbf R$ , $x_0$ is a limit point of $X$ , and $f : X \to \mathbf R$ is a function which is differentiable at $x_0$ . Let $Y \subseteq X$ be such that $x_0 \in Y$ , and $x_0$ is also a limit point of $Y$ . Prove that the restricted function $f|_Y : Y \to \mathbf R$ is also differentiable at $x_0$ , and has the same derivative as $f$ at $x_0$ . Explain why this does not contradict the discussion in Remark 10.1.2.

Hints

Use Definition 9.3.6.

How to think about the exercise

This is a simple exercise and is a matter of putting together the right definitions.

Model solution

We want to show that

$\displaystyle \lim_{y \to x_0; y \in Y\setminus\{x_0\}} \frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} = f'(x_0)$

To do this, we will return to the definition of a limit, Definition 9.3.6. To show the limit exists, we must first verify that $x_0$ is an adherent point of $Y\setminus\{x_0\}$ . But this is the case since $x_0$ is a limit point of $Y$ . Now let $\varepsilon > 0$ . We want to find some $\delta > 0$ such that for all $y \in Y\setminus\{x_0\}$ , if $|y-x_0| < \delta$ then $\left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon$ .

How can we find such a $\delta$ ? The only information we are given from which we could find a $\delta$ is the fact that $f$ is differentiable at $x_0$ . Since $f$ is differentiable at $x_0$ , we know that for our $\varepsilon > 0$ in particular, there exists some $\delta > 0$ such that for all $x \in X \setminus \{x_0\}$ , if $|x-x_0| < \delta$ then $\left|\frac{f(x) - f(x_0)}{x-x_0} - f'(x_0)\right| \leq \varepsilon$ .

So let’s use this $\delta > 0$ . Now that we have a $\delta$ , we must show that all $y \in Y\setminus\{x_0\}$ , if $|y-x_0| < \delta$ then $\left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon$ . So let $y \in Y\setminus\{x_0\}$ , and suppose $|y-x_0| < \delta$ . We must show that $\left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon$ .

Now $Y \subseteq X$ , so this means $Y \setminus \{x_0\} \subseteq X \setminus \{x_0\}$ . Thus we have $y \in X\setminus \{x_0\}$ .

We know from the differentiability condition for $f$ at $x_0$ that for all $x \in X \setminus \{x_0\}$ , if $|x-x_0| < \delta$ then $\left|\frac{f(x) - f(x_0)}{x-x_0} - f'(x_0)\right| \leq \varepsilon$ . Since our $y$ satisfies these conditions, we have $\left|\frac{f(y) - f(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon$ .

To complete the proof, recall how $f|_Y$ is defined. If $y' \in Y$ , then $f|_Y(y') := f(y')$ . Since $y,x_0 \in Y$ , we have $f|_Y(y) = f(y)$ and $f|_Y(x_0) = f(x_0)$ . Thus we have $\left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right|=\left|\frac{f(y) - f(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon$ .

This does not contradict the discussion in Remark 10.1.2 because in the example there, $3$ was not an adherent point of $[1,2]$ , so the limit was undefined. In contrast for this exercise we assumed that $x_0$ was an adherent point of $Y \setminus \{x_0\}$ (i.e. a limit point of $Y$ ). And so if we used the example in Remark 10.1.2, the proof above would fail at the point where we tried to verify that $x_0$ is an adherent point of $Y\setminus\{x_0\}$ .

The important point is that if $x_0$ were not a limit point of $Y$ , then we could always pick $\delta > 0$ small enough so that the expression $\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0}$ is undefined when $y$ is further restricted to the set $\{y \in Y \setminus \{x_0\} : |y - x_0| < \delta\}$ . This would mean that the limit is always defined no matter what $L$ we use (since the implication in the definition of limit would be vacuous), so that in effect the derivative would equal any number possible. This would make the notation $f|_Y'(x_0)$ ambiguous, and this concept is useless anyway, which is why we choose to not define the limit in this case.