3
$\begingroup$

Consider a sequence of functions $f_n(x)$, which converges to $f(x)$ pointwise on the interval $(0,1)$. The functions $f_n(x)$ and the limit function $f(x)$ are differentiable, with derivatives $f'_n(x)$ and $f'(x)$ respectively.

Assume that the sequence $f'_n(x)$ converges to a function pointwise. Because convergence of the derivatives is not necessarily uniform, the usual theorem for term-by-term differentiability cannot be applied to give $\lim_{n \rightarrow \infty} f'_n(x) = f'(x)$. However,

If it is known that both $f_n(x)$ and $x f'_n(x)$ converge uniformly on $[0,1]$, does that imply that $\lim_{n \rightarrow \infty} f'_n(x) = f'(x)$ on $(0,1)$?

I think the answer is yes. But it strikes me a bit that I get the desired result side-stepping the need for uniform convergence of $f'_n(x)$.

So I'll post my proof and please let me know if there's any mistake in it. Or if the result is well-known, can you point me to a relevant theorem?

$\endgroup$
2
  • $\begingroup$ How about the case that we don't have such uniform convergence? Is there some examples that $f'n$ converges other than $f$, or $f'_n$ diverges for some $x$ even though $f'(x)$ exists? $\endgroup$ Commented Nov 21, 2023 at 7:40
  • $\begingroup$ @qkqh I don't understand your question, sorry. Maybe post a separate question, possibly with reference to this one $\endgroup$ Commented Nov 21, 2023 at 10:25

1 Answer 1

Reset to default
3
$\begingroup$

Yes, the implication holds

Proof

Since $f_n(x)$ converges to $f(x)$ uniformly on the compact interval $[0,1]$, $x f_n(x)$ converges to $x f(x)$ uniformly on that interval.

The derivative of $xf_n(x)$ is $f_n(x) + x f'_n(x)$, which by the assumptions converges uniformly on $[0,1]$. Therefore, the referred theorem can be applied to the sequence of functions $x f_n(x)$. As a result, the derivatives of these functions converge to the derivative of $x f(x)$; that is, $f_n(x) + x f'_n(x)$ converges to $f(x) + x f'(x)$. This implies that $x f'_n(x)$ converges to $x f'(x)$. For $x \in (0,1)$, dividing by $x$ shows that $f'_n(x)$ converges to $f'(x)$.

$\endgroup$
1
  • $\begingroup$ How about the case that we don't have such uniform convergence? Is there some examples that $f'n$ converges other than $f$, or $f'_n$ diverges for some $x$ even though $f'(x)$ exists? $\endgroup$ Commented Nov 21, 2023 at 7:40

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.