Erroneous Proof that the Derivative is Continuous

Question

I have written the following which argues that for any arbitrary sequence of points such that $x_n \to c, x_n\neq c$ we must also have $f'(x_n)\to f'(c)$ which would imply that $f'(x)\to f'(c)$ as $x\to c$ , i.e., the derivative is continuous. However, I can't see where the error in my argument is.

Let $f:[a,b] \to \mathbb{R}$ be a differentiable function and $c\in [a,b]$. Fix $\epsilon > 0$. By assumption, for $c\in [a,b]$, $f'(c)$ exists so put $\delta_1 > 0$ so small that whenever $0<\lvert x -c \rvert < \delta_1$ we must have $$ \left \lvert \frac{f(x)-f(c)}{x-c}-f'(c) \right \rvert < \frac{\epsilon}{2} $$ $[a,b]$ is closed so there exists a sequence such that $x_n \to c$ and $x_n \neq c$. By assumption, $f'(x_n)$ exists so there exists a $\delta_2 >0$ such that whenever $0<\lvert x-x_n \rvert < \delta_2$ we must have $$ \left \lvert f'(x_n)-\frac{f(x)-f(x_n)}{x-x_n} \right \rvert < \frac{\epsilon}{2} $$ Now pick an integer $N$ so large that for $n\geq N$ we must have $0<\lvert x_n - c\rvert < \mathrm{min}(\delta_1,\delta_2)$. It follows that $$ \left \lvert \frac{f(x_n)-f(c)}{x_n-c}-f'(c)\right \rvert < \frac{\epsilon}{2} \quad \textbf{and} \quad \left \lvert f'(x_n)-\frac{f(x_n)-f(c)}{x_n-c}\right \rvert < \frac{\epsilon}{2} $$ Applying the triangle inequality, we see that $\lvert f'(x_n)-f'(c) \rvert < \epsilon$ and we are done.

Answer 1

Let me write it up so the question is not unanswered...

Note that your $\delta_2$ depends on $x_n$. So we should really write $\delta_2(n)$. That is for a fixed $n$, $\delta_2(n)$ is a positive real such that if $0\lt |x-x_n|\lt \delta_2(n)$, then $\left|\frac{f(x)-f(x_n)}{x-x_n} - f'(x_n)\right| \lt \epsilon/2$.

Then $\delta=\min(\delta_1,\delta_2(n))$ guarantees that $|f'(c)-f'(x_n)|\lt \epsilon$ for that fixed $n$, not for an arbitrary $n$.

So in order to ensure that this inequality holds for all $n\geq N$, you would need to find a $\delta$ that satisfies $\delta\leq \inf(\{\delta_1\}\cup\{\delta_2(n)\mid n\geq N\})$. Unfortunately, that infimum could be equal to $0$, which means that you have no guarantee that you can pick a $\delta\gt 0$ that ensures both inequalities you have at the end for all $n\geq N$.

Answer 2

The problem is that your proof is not written well enough, so the mistake may be hard to find.

Something better would have been the following.

1. Blah blah.

...

23. For every $n$, there exists $\delta_2$ such that blah blah blah.

24. Let $\delta_2$ such that for every $n$, blah blah blah (such a $\delta_2$ exists from 23)).

...

53. So $f’$ is continuous.

In this proof, it is now quite obvious to spot the « mistake »: while $\forall n, \ \exists \delta_2$ is proved, you use the unproved assertion $\exists \delta_2, \ \forall n$.