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This question arose while studying chapter 1 of Reed & Simon's book 'Methods of Modern Mathematical Physics'. It is about exercise 1.10 in particular. In this exercise, the following is asked (I reproduce all the relevant information given in the chapter):

Let $x_0,\dots,x_n$ be a partition of the interval $[a,b]$, $x_0=a, x_n=b$. Let $\chi_i(x)$ be the characteristic function of $[x_{i-1},x_i)$ except for $\chi_n(x)$ which is the char. function of $[x_{n-1},x_n]$. A function on $[a,b]$ of the form $\sum_{i=1}^n s_i\chi_i(x)$ is called a step function. The set of all step functions for all possible finite partitions of $[a,b]$ is a normed linear space with the norm $$\left\|\sum_{i=1}^n s_i\chi_i(x)\right\|_\infty=\sup_{x\ \in\ [a,b]}\left|\sum_{i=1}^n s_i\chi_i(x)\right|=\max_{i=1,\dots,n}\{s_i\}$$
We denote this space by $S[a,b]$.

Exercise 10: Prove that $S[a,b]$ is dense in the norm $\|\cdot\|_\infty$ in $PC[a,b]$, the space of all possible bounded piecewise continuous functions on $[a,b]$.

The definition of 'dense' was given earlier in the book:

Definition: A set $B$ in a metric space $M$ is called dense if every $x\in M$ is a limit of elements of $B$.

I am quite confused about what one is supposed to do to solve the exercise: Is it asking me to show that the $\|\cdot\|_\infty$-norm of any function in $PC[a,b]$ is the limit of the $\|\cdot\|_\infty$-norm of some function in $S[a,b]$? That statement seems trivial since, with these definitions, the $\|\cdot\|_\infty$-norm of a bounded function is just its supremum, so any step function with a suitably chosen highest $s_i$ has the same $\|\cdot\|_\infty$-norm (or am I missing something here?).

Is it something else (plausibly something involving the metric naturally induced by this norm: $d(f,g)=\|f-g\|_\infty$?) that I would have to show instead?

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    $\begingroup$ Dense in a norm generally means dense with respect to the topology induced by the metric induced by that norm. (With your definition of density, you'd just show that it's dense using the metric induced by this norm) $\endgroup$
    – qaphla
    Jul 7, 2014 at 23:44
  • $\begingroup$ @qaphla Thank you for your response. However, my main confusion is: What exactly does it mean to show that it's dense using the metric induced by this norm? Would it be possible for you to rephrase the question (perhaps in terms of existential/universal quantifiers to maximize clarity)? $\endgroup$
    – Danu
    Jul 7, 2014 at 23:53

1 Answer 1

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Phrases like "dense with respect to a norm" should be interpreted as "dense with respect to the metric (or topology) induced by the norm". (You should be familiar with metric space and point-set topology before trying to read Reed and Simon.)

So one way of rephrasing the question is as follows:

Show that for each $f \in PC[a,b]$ there exists a sequence $\{f_n\} \subset S[a,b]$ such that $\lim_{n \to \infty} \|f_n - f\|_\infty = 0$.

This is not the same as the statement:

[WRONG] Show that for each $f \in PC[a,b]$ there exists a sequence $\{f_n\} \subset S[a,b]$ such that $\lim_{n \to \infty} \|f_n\|_\infty = \|f\|_\infty$.

The second statement is much weaker than the first.

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