Analysis III TT23, Differentiation and limits
Flashcards
@State the theorem that lets you do the following
\[\lim _ {n \to \infty} f' _ n = \left(\lim _ {n \to \infty} f _ n\right)'\]
?
Suppose that $f _ n : [a, b] \to \mathbb R$ is a sequence of functions such that
- $f _ n$ is continuously differentiable on $(a, b)$
- $f _ n$ converges pointwise to some function $f$ on $[a, b]$.
- $f’ _ n$ converges uniformly to some bounded function $g$ on $[a, b]$
then $f$ is differentiable and
\[\lim _ {n \to \infty} f' _ n = \left(\lim _ {n \to \infty} f _ n\right)'\]@State the theorem that lets you do the following
\[\left(\sum _ i \phi _ i\right)' = \sum _ i \phi _ i'\]
?
Suppose that $\phi _ i : [a, b] \to \mathbb R$ is a sequence of functions continuously differentiable on $(a, b)$ with $\sum _ i \phi _ i$ converging pointwise. Suppose that $ \vert \phi _ i'(x) \vert \le M _ i$ for all $x \in (a, b)$ where $\sum _ i M _ i < \infty$. Then $\sum \phi _ i$ is differentiable and
\[\left(\sum _ i \phi _ i\right)' = \sum _ i \phi _ i'\]Can you give an example of a sequence of functions $f _ n$ all differentiable that converge uniformly but where the derivative of the limit is not equal to the pointwise limit of the derivatives?
Proofs
@Prove that if $f _ n : [a, b] \to \mathbb R$ is a sequence of functions such that
- $f _ n$ is continuously differentiable on $(a, b)$
- $f _ n$ converges pointwise to some function $f$ on $[a, b]$.
- $f’ _ n$ converges uniformly to some bounded function $g$ on $[a, b]$
then $f$ is differentiable and
\[\lim _ {n \to \infty} f' _ n = \left(\lim _ {n \to \infty} f _ n\right)'\]
Todo.