Space of Piecewise Linear Functions on Closed Interval is Dense in Space of Continuous Functions on Closed Interval
Theorem
Let $I = \closedint a b$.
Let $\map \CC I$ be the set of continuous functions on $I$.
Let $\map {\mathrm {PL} } I$ be the set of piecewise linear functions on $I$.
Let $d$ be the metric induced by the supremum norm.
Then $\map {\mathrm {PL} } I$ is dense in $\struct {\map \CC I, d}$.
Proof
Let $f \in \map \CC I$.
Let $\epsilon \in \R_{>0}$ be a real number.
From Open Ball Characterization of Denseness:
- it suffices to find a $p \in \map {\mathrm {PL} } I$ such that $p$ is contained in the open ball $\map {B_\epsilon} f$.
From Continuous Function on Closed Real Interval is Uniformly Continuous:
- $f$ is uniformly continuous on $I$.
That is:
- there exists a $\delta > 0$ such that for all $x, y \in I$ with $\size {x - y} < \delta$ we have $\size {\map f x - \map f y} < \epsilon/3$.
Let:
- $P = \{a_0 = a, a_1, a_2, \ldots, a_n = b\}$
be a finite subdivision of $I$, with:
- $\size {a_{i + 1} - a_i} < \delta$
for each $i$.
Let $p \in \map {\mathrm {PL} } I$ be such that:
- $\map p {a_i} = \map f {a_i}$
for each $i$, with $p$ continuous.
We can explicitly construct such a $p$ by connecting $\tuple {a_i, \map f {a_i} }$ to $\tuple {a_{i + 1}, \map f {a_{i + 1} } }$ with a straight line segment for each $i$.
Fix $x \in I$.
Note that there exists precisely one $i$ such that $a_i \le x \le a_{i + 1}$, fix this $i$.
We then have:
\(\ds \size {\map p x - \map f {a_i} }\) | \(=\) | \(\ds \size {\map p x - \map p {a_i} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {\map p {a_{i + 1} } - \map p {a_i} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \size {\map f {a_{i + 1} } - \map f {a_i} }\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon/3\) |
since $\size {a_{i + 1} - a_i} < \delta$.
Since $\size {x - a_i} < \size {a_{i + 1} - a_i} < \delta$, we also have:
- $\size {\map f x - \map f {a_i} } < \epsilon/3$
So:
\(\ds \size {\map p x - \map f x}\) | \(\le\) | \(\ds \size {\map p x - \map p {a_i} } + \size {\map f x - \map f {a_i} }\) | Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds 2 \epsilon/3\) |
Note that $x \in I$ was arbitrary, so:
\(\ds \map d {f, p}\) | \(=\) | \(\ds \norm {f - p}_\infty\) | Definition of Metric Induced by Norm | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup_{x \mathop \in I} \size {\map f x - \map p x}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \epsilon/3\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) |
so $p \in \map {B_\epsilon} f$.
$\blacksquare$