A corner to share my understanding of various mathematical topics, in a fun way (but not always).
Convergence is a topic which starts to appear right from the beginning of a maths degree. A good way to understand this concept is to categorise it by what is converging:
A sequence $x_n$ converges to $x$ if for every $\epsilon$, there exists $N$ such that
\(n \ge N \implies |x_n - x| < \epsilon\) .
\(x_n = \frac{1}{n} \to 0\) .
\(x_n = (-1)^n\) , does not converge.
A sequence $x_n$ is Cauchy if for every $\epsilon > 0$, there exists $N$ such that
\(m, n \ge N \implies |x_n - x_m| < \epsilon\) .
\(x_n = \frac{1}{n} \to 0\) .
In fact, in $\mathbb{R}, \mathbb{C}$, Cauchy $\leftrightarrow$ convergent.
If $x_n$ is monotone and bounded, then it converges.
\(x_n = 1-\frac{1}{n},\) is increasing and bounded above by 1, so $x_n \to 1$.
Let $f_n: E \to \mathbb{R}$, f: E \to \mathbb{R}$.
\(f_n \to f\) , pointwise on $E$ means:
for every fixed $x \in E$,
\(f_n(x) \to f(x)\) .
Equivalently,
\(\forall x \in E, \forall \epsilon > 0, \exists N = N(x, \epsilon) \, \text{such that} \, n \ge N \implies |f_n(x) - f(x)| < \epsilon\) .
Example: On $[0, 1], \, f_n(x)=x^n.$ Then
\(f_n(x) \to f(x) = \begin{cases} 0, \quad 0 \le x < 1 \\ 1, \quad x = 1 \end{cases}\) .
So pointwise convergence holds.
Note that each $f_n(x) = x^n$ is continuous, but the limit $f$ is not continuous at 1. So pointwise convergence does not preserve continuity.
\(f_n \to f\) , uniform on $E$ means:
\(\forall \epsilon > 0, \exists N = N(\epsilon) \, \text{such that} \, n \ge N \implies |f_n(x) - f(x)| < \epsilon, \, \forall x \in E\) .
Equivalent form:
\(\sup_{x\in E} |f_n(x) \to f(x)| \to 0.\) .
Example:
On $[0, 1], \, f_n(x)= \frac{x}{n}.$ Then
\(\sup_{x\in[0, 1]} |f_n(x)| = \frac{1}{n} \to 0,\) so $f_n \to 0$, uniformly.
Standard counterexample: $f_n(x) = x^n$ converges pointwise, but not uniformly, because \(\sup_{x\in[0, 1]} |x^n - f(x)| = 1, \quad \forall n\) .
Note that uniform convergence preserves continuity.
\(f_n \to f\) , locally uniform on $U$,
means uniform convergence on every compact subset $K \subset U$.
Example:
\(f_n(x) = \frac{x}{n}\) ,
converges locally uniformly on $\mathbb{R}$, in fact, uniformly on every bounded interval.
Counterexample to global uniformity:
\(f_n(x) = \frac{x}{n}, \quad x \in \mathbb{R}\) , does not converge uniformly on all of $\mathbb{R}$, because
\(\sup_{x\in\mathbb{R}} \left| \frac{x}{n} \right| = \infty\) .
In a metric space $(X, d), x_n \to x$, means
\(d(x_n, x) \to 0\) .
Example:
In $C([0, 1])$ with the sup metric
\(d(f, g) = \sup_{x\in[0, 1]} |f(x) - g(x)|\) ,
metric convergence is exactly uniform convergence.
In $L^p$, metric convergence is $L^p$-convergence.
In a normed space $(X, |\cdot|$,
\[x_n \to x \quad \text{strongly (or in norm)}\]means,
\(\| x_n - x\| \to 0\) .
Example:
\(\|f_n - f\|_\infty \to 0\) is uniform convergence.
Let $X$ be a normed space and $X^*$ be its dual. Then
\(x_n \rightharpoonup x\) , means:
for every continuous linear function $\varphi\in X^*$
\(\varphi(x_n) \to \varphi(x)\) .
Example: In a Hilbert space $l^2$, the standard basis $e_n \rightharpoonup 0$ weakly:
for any $y = (y_k) \in l^2$,
\(\langle y, e_n \rangle = y_n \to 0\)
.
But $e_n \not \to 0$, because $| e_n | = 1$.
Note that convergence in norm implies weak convergence, but not vice versa in infinite-dimensional spaces.
If $\varphi_n^* \in X^*$ then
\(\varphi_n^* \rightharpoonup \varphi^*\) ,
means
$\forall x \in X$,
\(\varphi_n^*(x) \to \varphi^*(x)\) .
Example:
In $l^\infty = (l^1)^* $, weak$-^*$ convergence means pointwise convergence on $l^1$-pairings.
\(f_n(x) \to f(x) \quad \text{for almost every $x$}\) ,
meaning except on a set of measure zero.
Example:
\(f_n(x) = x^n \quad \text{on} \quad [0, 1]\) , converges a.e. to $0$, because only $x=1$, the limit is 1, and ${1}$ has measure zero.
Counterexample: a.e. convergence does not imply $L^1$ convergence.
Take
\(f_n(x) = n \mathbf{1}_{(0, 1/n)}(x) \quad [0, 1]\) .
Then, $f_n(x) \to 0$, $\forall x > 0$, hence, a. e. to 0. But
\(\| f_n\|_1 = \int_0^1 n \mathbf{1}_{(0, 1/n)}(x) dx = 1\) ,
so not $L^1$-convergence to 1.
\(f_n \to f \quad \text{in measure}\) , means,
$\forall \epsilon >0$
\(\mu(x\{x: |f_n(x) - f(x)|> \epsilon \}) \to 0\) .
Example:
If $f_n \to f$ in $L^p$, then $f_n \to f$ in measure.
Counterexample:
Convergence in measure does not imply a.e. convergence for the whole sequence in general.
Standard example:
The ‘typewriter sequence’ on [0,1], where intervals sweep across dyadic subintervals. It converges to 0 in measure but not pointwise a.e.
For $1\le p < \infty$,
\(f_n \to f, \quad \text{in} \quad L^p\) means
\(\| f_n - f\|_p = \left( |f_n - f|^p\right)^{1/p} \to 0\) .
For $L^\infty\(,\) f_n \to f, \quad \text{in} \quad L^\infty $$ means
\(\| f_n - f\|_\infty = \text{esssup} |f_n - f| \to 0\) .
Example:
\(f_n(x) = \frac{1}{n} \to 0 \quad \text{in every} L^p([0, 1])\) .
Counterexample:
\(f_n(x) = n \mathbf{1}_{(0, 1/n)}(x)\) ,
converges a.e. to 0, but not in $L^1$.
Another example:
\(f_n(x) = n \mathbf{1}_{(0, 1/n)}(x)\) ,
then $f_n \to 0$ a.e. and in $L^p$ for every finite $p$, because
\(\| f_n\|_p = (1/n)^{1/p} \to 0\) .