# approximation of 'any' bounded continuous function using bounded continuous functions with compact support

Suppose that $$\phi$$ is a bounded, continuous function with compact support $$I$$ (i.e. a bounded interval), then given any $$\epsilon > 0$$, there exists a simple function $$\phi_{\epsilon}$$ s.t.

$$\text{sup}_{x\in I} |\phi(x)-\phi_{\epsilon}(x)| \leq \epsilon \quad \quad (\text{eq. 1})$$

which I guess should mean that there exists an increasing sequence of simple functions, $$\phi_{\epsilon}(x)$$ which converges uniformly to $$\phi(x)$$ due to its compact support...

what I don't understand is the passage from bounded, continuous functions with compact support to 'any' kind of bounded continuous function. The text says:

"any bounded, continuous function $$\phi^*$$ (not necessarily having compact support) may be approximated by an increasing sequence of bounded, continuous functions $$\phi$$ with compact support, i.e. $$\phi_M(x) = \phi^*\textbf{1}_{[-M,M]}$$ will converge to $$\phi^*(x)$$ as $$M \rightarrow \infty$$"

I got confused by this part.. If $$M \rightarrow \infty$$, shouldn't $$\phi_M(x)$$ have unbounded (non-compact) support at the limit, in this case, how will we retain the property in $$\textbf{eq. 1}$$?

• Everything made sense up to your last sentence. The claim being made is that, for any particular value of $M$, (for example, $M = 10000$), $\phi_M$ has compact support. That is, each function in the sequence $\phi_1, \phi_2, \phi_3, ...$ has compact support. – Jake Mirra Oct 22 at 2:41
• Or perhaps you mean, how do you get approximation by simple functions. Well, if $\phi_M$ is close to $\phi$ and $\phi_\epsilon$ is close to $\phi_M$, then by the triangle inequality, $\phi_\epsilon$ is close to $\phi$. – Jake Mirra Oct 22 at 2:46

It is possible to have $$\phi_\varepsilon \rightarrow \phi^*$$, but it is not possible (in general) for $$\phi_\varepsilon$$ to satisfy the uniform convergence estimate (1). For example, $$\phi^*(x) \equiv 1$$ for $$x \in \mathbb{R}$$ clearly cannot be uniformly approximated by a simple function with compact support. But it is also clear that $$\mathbb{1}_{[-M,M]}(x) \rightarrow \phi^*(x)$$ pointwise on $$\mathbb{R}$$.
To prove the pointwise estimate, fix $$\varepsilon > 0$$ and $$\phi_M$$ satisfying $$|\phi_M(x) - \phi^*(x)| < \varepsilon/2$$ for all $$x \in [-M,M]$$, and then, once this $$\phi_M$$ is chosen, find a simple function $$\phi_\varepsilon$$ satisfying $$|\phi_M(x) - \phi_\varepsilon(x) | < \varepsilon / 2$$ for all $$x \in [-M, M]$$. By the triangle inequality we can combine these two to get $$|\phi^*(x) - \phi_\varepsilon(x) | < \varepsilon$$ for all $$x \in [-M,M]$$.