differentiable and sequences of functions

Question

Let $f_n:(a,b)\rightarrow \mathbb{R}$ be differentiable and $f_n\rightarrow f$.

Can $f$ be differentiated?

Would it be $f'(x_0)=\lim_{n \to \infty}f_n'(x_0)$ for $x_0\in (a,b)$ ?

Answer 1

Can $f$ be differentiated?

Not necessarily. Consider $f_n(x) = \sqrt{x^2 + \frac{1}{n}}$ on $(-a, a)$. $f_n(x) \to f (x) = |x|$ which is not differentiable at $0$.

Would it be $f'(x_0)= \lim_{n\to\infty} f_n'(x_0)$ for $x_0 \in (a,b)$ ?

Not necessarily. Let $f_n(x) = \frac{\sin(nx)}{\sqrt{n}}, n \ge 1$ for $x\in(-a, a)$ where $a > 0$.

$f_n \to f = 0$.

So, $f'(x) = 0$ for all $x\in(-a, a)$. $f_n'(x) = \sqrt{n}\cos(nx)$. However, $f_n'(0) = \sqrt{n} \to \infty$ but $f'(0) = 0$.

Answer 2

For the supremum norm, this is not true in general. You are trying to change the order of limits (in the sequence and in the derivative) which generally changes the result. For example, consider a sequence of differentiable functions that tend towards an indicator function of a set. The limit is not differentiable. If you have studied Fourier series you can generate similar examples to any integrable function.

If you define a different norm, say where you take the supremum plus the supremum of the first derivative, then you would have convergence, and there are other examples of norms where you do get convergence to some notion of a derivative.

Answer 3

No.

You need a stronger assumption that ${f_n'}$ converges uniformly on (a b).

Neither ${f_n}$ --> f pointwise on (a b), nor ${f_n}$ --> f uniformly on (a b) is good enough (See Example 7.5 a counter example) for the result to hold.

See W. Rudin, PMA, Theorem 7.17 for under what condition, the result is true.