Analysis II HT23, Uniform convergence
Flashcards
@Define what it means for $(f _ n)$ to converge pointwise to $f : E \to \mathbb R$.
What does the sequence of functions $f _ n(x) = x^n$ where $f _ n : [0, 1] \to \mathbb{R}$ converge (pointwise) to?
If $f _ n$ is continuous for every $n$, and $f _ n$ converges pointwise to $f$, then is $f$ continuous?
Not necessarily.
Let $(f _ n)$ be a sequence of functions where $f _ n : E \to \mathbb R$. @Define what it means for $(f _ n)$ to converge uniformly to $f$.
In general, if every $f _ n$ in $(f _ n)$ is continuous and $(f _ n)$ converges pointwise to $f$, then $f$ is not necessarily continuous. But what stronger condition allows you to say $f$ will definitely be continuous if $f _ n$ is?
Uniform convergence.
Uniform continuity and uniform convergence both are stronger conditions than ordinary continuity and ordinary pointwise convergence. What’s the similarity between them?
They require showing that some value of $\delta$ or $N$ works for all points.
What notation is used for a sequence of functions $(f _ n)$ converging uniformly to $f$?
Let $(f _ n)$ be a sequence of functions where $f _ n : E \to \mathbb R$.
The definition of uniform convergence to $f$,
\[\forall \varepsilon> 0 \colon \exists N \in \mathbb N \text{ s.t. } \forall n>N \colon \forall x \in E \colon \vert f _ n(x) - f(x) \vert < \varepsilon\]
is quite complicated and hard to use in practice. Can you state the equivalent definition in terms of supremums?
$(f _ n)$ converges uniformly to $f$ if and only if for sufficiently large $n$, $\{ \vert f _ n(x) - f(x) \vert : x \in E\}$ is bounded, and
\[s _ n = \sup _ {x \in E} \vert f _ n(x) - f(x) \vert \to 0 \text{ as } n \to \infty\]@State the Cauchy criterion for uniform convergence.
Suppose $f _ n : E \to \mathbb R$, then $f _ n$ converges uniformly if and only if
\[\forall \varepsilon > 0 \text { } \exists N \in \mathbb N \text{ s.t. } \forall m > n > N \text{ } \forall x \in E \text{ } \vert f _ n(x) - f _ m(x) \vert < \varepsilon\]When is the Cauchy criterion for uniform convergence
\[\forall \varepsilon > 0 \text { } \exists N \in \mathbb N \text{ s.t. } \forall m > n > N \text{ } \forall x \in E \text{ } \vert f _ n(x) - f _ m(x) \vert < \varepsilon\]
useful?
When we don’t have an expression for the function the series is converging to, i.e. when defining functions via power series.
@State the (slightly simpler) Cauchy criterion for uniform convergence for a sum $f _ n = \sum _ {k=1}^n u _ k$ of functions.
@State Weierstrass’ M-test for determining if a sum of functions converges uniformly.
Suppose $u _ k : E \to \mathbb R$ and $M _ k \in \mathbb R$ where $M _ k$ are constants, $\forall x \in E$, $ \vert u _ k(x) \vert \le M _ k$, and $\sum _ {k=1}^\infty M _ k$ converges, then
\[\sum u _ k\]converges uniformly.
What useful theorem relates uniform convergence and power series?
If $\sum c _ k x^k$ has radius of convergence $R$ and if $ \vert \rho \vert < R$, then $\sum c _ k x^k$ converges uniformly on $\{x : \vert x \vert \le \rho\}$.
There is a theorem that states
If $\sum c _ k x^k$ has radius of convergence $R$ and if $\rho < R$, then $\sum c _ k x^k$ converges uniformly on $\{x : \vert x \vert \le \rho\}$.
Why is the condition about $\rho$ required, rather than $ \vert x \vert < R$?
Otherwise you could get a one-sided limit at the boundary.
If a sum of differentiable functions converges uniformly to $f$, does that mean that $f$ is also differentiable, like is the case with continuity?
No.
When proving that power series converge uniformly within their radius of convergence, i.e.
If $\sum c _ k x^k$ has radius of convergence $R$ and if $\rho < R$, then $\sum c _ k x^k$ converges uniformly on $\{x : \vert x \vert \le \rho\}$.
you use the $M$-test. How do you define $M _ k$, and can you verify that $u _ k(x) \le M _ k$ and that $\sum M _ k$ converges?
Define $M _ k = \vert c _ k \rho^k \vert $. Note $u _ k(x) = \vert c _ k x^k \vert \le M _ k$, and since $\rho$ is within the radius of convergence, $\sum \vert c _ k \rho^k \vert = \sum M _ k$ converges.
When proving Weierstrass’ M-test, i.e.
Suppose $u _ k : E \to \mathbb R$ and $M _ k \in \mathbb R$ where $M _ k$ are constants, $\forall x \in E$, $ \vert u _ k(x) \vert \le M _ k$, and $\sum _ {k=1}^\infty M _ k$ converges, then
\[ \sum u _ k(x) \]
converges uniformly.
what do you note to get started that the rest of the proof follows from almost immediately?
$\sum^{\infty} _ {k=1} M _ k$ converges implies the Cauchy criterion, i.e. that
\[ \vert M _ {n+1} + \cdots + M _ {m} \vert = M _ {n+1} + \cdots + M _ m < \varepsilon\]Can you give an example of a power series that is uniformly convergent within any bounded interval inside the interval of convergence, but not uniformly convergent on the whole interval?
Consider
\[f _ n(x) = \frac{nx}{n^2x^2 + 1}\]
one way to show that this isn’t uniformly convergent on $[0, 1]$ is to use derivatives. What could you quickly note instead?
Proofs
@Prove the following.
Let $(f _ n)$ be a sequence of continuous functions on $E$ which converges uniformly to $f$ on $E$. Then if each $f _ n$ is continuous at $p \in E$, then $f$ is continuous at $p$.
Todo.
@Prove the equivalence between uniform convergence and the following supremum condition:
$(f _ n)$ converges uniformly to $f$ if and only if for sufficiently large $n$, $\{ \vert f _ n(x) - f(x) \vert : x \in E\}$ is bounded, and
\[s _ n \coloneqq \sup _ {x \in E} \vert f _ n(x) - f(x) \vert \to 0 \text{ as } n \to \infty\]
Todo.
@Prove the Cauchy criterion for uniform convergence:
Suppose $f _ n : E \to \mathbb R$, then $f _ n$ converges uniformly if and only if
\[\forall \varepsilon > 0 \text { } \exists N \in \mathbb N \text{ s.t. } \forall m > n > N \text{ } \forall x \in E \text{ } \vert f _ n(x) - f _ m(x) \vert < \varepsilon\]
Todo.
@Prove Weierstrass’ M-test:
Suppose $u _ k : E \to \mathbb R$ and $M _ k \in \mathbb R$ where $M _ k$ are constants, $\forall x \in E$, $ \vert u _ k(x) \vert \le M _ k$, and $\sum _ {k=1}^\infty M _ k$ converges, then
\[\sum u _ k(x)\]
converges uniformly.
Todo.
@Prove the uniform convergence of power series within their radius of convergence:
If $\sum c _ k x^k$ has radius of convergence $R$ and if $\rho < R$, then $\sum c _ k x^k$ converges uniformly on $\{x : \vert x \vert \le \rho\}$.
Todo.