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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 $\textbf{eq. 1}$?

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  • $\begingroup$ 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. $\endgroup$
    – Jake Mirra
    Commented Oct 22, 2019 at 2:41
  • $\begingroup$ 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 $. $\endgroup$
    – Jake Mirra
    Commented Oct 22, 2019 at 2:46

1 Answer 1

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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] $.

Hope that's the help you were looking for.

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