If \(\mathbf{a}\) and \(\mathbf{b}\) are two finite-dimensional complex (or real) vectors, the distance between them is precisely \(\| \mathbf{a}-\mathbf{b}\|\).
This suggests that a good notion of convergence of a sequence of a sequence of vectors is:
Definition (Norm convergence of vectors)
Let \((\mathbf{a}_n)_{n=1}^\infty\) be a sequence of vectors and \(\mathbf{a}\) a fixed vector in \(\mathbb{C}^k\). Then the sequence converges to \(\mathbf{a}\) in norm, written \[ \lim_{n\to \infty} \mathbf{a}_n = \mathbf{a}, \] if \[ \lim_{n\to \infty} \|\mathbf{a}_n - \mathbf{a}\|=0. \]
Definition (Norm convergence of functions)
Let \((f_n)_{n=1}^\infty\) be a sequence of functions and \(f\) a fixed function in \(PC(a,b)\). Then the sequence converges to \(f\) in norm, written \[ \lim_{n\to \infty} f_n = f, \] if \[ \lim_{n\to \infty} \|f_n - f\|=0. \]
Pointwise and norm convergence are completely different modes of convergence.
Given the sequence of functions \[ f_n(x) = \begin{cases} 1 & : 0 \leq x \leq 1/n, \\ 0 & : 1/n < x \leq 0, \end{cases} \] on \([0,1]\), prove that \(f_n\to 0\) in norm, but that the \(f_n\)’s do not converge to \(0\) pointwise.
Conversely, given the sequence of functions \[ g_n(x) = \begin{cases} n & : 0 < x < 1/n, \\ 0 & : x=0, \, \text{or} \, 1/n \leq x \leq 1, \end{cases} \] on \([0,1]\), prove that \(g_n\to 0\) pointwise, but that the \(g_n\)’s do not converge to \(0\) in norm.
However! We do have the following simple result.
To state it, recall that a sequence of functions \(f_n\to f\) uniformly (on \([a,b]\)) if there is a sequence of real constants \((M_n)_{n=1}^\infty\) such that \[ |f_n(x) - f(x)| \leq M_n \] for all \(x\in [a,b]\) and \(M_n \to 0\).
Theorem 3.2
If \(f_n\to f\) uniformly on \([a,b]\), then \(f_n\to f\) in norm.
The main technique in calculus and analysis is to construct objects by “passing to the limit.” This is how derivatives are constructed, for example, as well as integrals.
This only works if the limit actually exists!
In this respect, the space of functions \(PC(a,b)\) is not ideal. There are sequences of functions in this space that “look like” they should converge, but don’t.
This is captured by saying that \(PC(a,b)\) is not complete.
To explain what this means, we need the following:
Definition
A sequence of functions \((f_n)_{n=1}^\infty\) is called a Cauchy sequence if \[ \|f_m - f_n\| \to 0 \] as \(m,n\to \infty\).
The same definition applies to other vector spaces with norms, like \(\mathbb{C}^k\).
Intuitively, this means that the functions in the sequence get closer and closer together the further one goes out in the sequence.
Definition
A vector space with a norm (like \(\mathbb{C}^k\), or \(PC(a,b)\)) is called complete if all Cauchy sequences converge (in norm).
By examining the sequence of functions \[ f_n(x) = \begin{cases} 0 & : 0 \leq x \leq 1/n, \\ x^{-1/4} & : 1/n < x \leq 1, \end{cases} \] on \([0,1]\), prove that \(PC(0,1)\) is not complete.