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How can one show that the space $C[a,b]$ (space of continuous functions on the interval a to b) is dense in $\check{C}[a,b]$ (space of piecewise continuous functions from a to b) under the norm induced by : $(a,b,) = \int a\overline{b}$, but not under $||f||_\infty = \sup |f(x)|$ ?


Let $(X,T)$ be a topological space, $(N,d)$ a metric space,

Definition of neighbourhood: In this case the $\epsilon$-neighbourhood: $$U_{\epsilon}(x):= \{y\in N | d(x,y)<\epsilon \}$$

Definition of dense: A subset M is then dense in X , if every neighbourhood in X contains a point of M


What I have tried:

If $C[a,b]$ lies dense in in $\check{C}[a,b]$, that means that for every $f \in C[a,b]$ and every $\epsilon > 0$ there must exist a $c\in \check{C}[a,b]$ with : $$(f,c) = \int f\overline{c} < \epsilon$$

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1 Answer 1

up vote 1 down vote accepted

Generally, $(f,g) \to \int f \bar{g}$ is an inner-product which induces a norm defined by $\|f\| := \sqrt{(f,f)}$. Actually, every inner product space gives rise to a normed space under this definition. We refer to this particular norm as the L 2 norm and often write $\|f\|_2$ or $\|f\|_{L^2}$, or even sometimes $\|f\|_{L^2([a,b])}$ if we want to emphasize where we're considering this norm.

So what you need to show is that given any piecewise continuous $f$, there's a continuous $g$ so that $\int (f - g)^2 < \epsilon$. This isn't bad. Given such an $f$, there are at most a finite number $x_1 < ... <x_N$ points of discontinuity. Set $g = f$ for the majority of these intervals and use the remaining room to interpolate between endpoints to keep $g$ continuous.

To show that these functions aren't dense under the sup norm, take a piecewise continuous function that has a large jump. For an appropriately picked $\epsilon > 0$, any continuous $f$ will have to differ by at least $\epsilon > 0$ somewhere near this jump.

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