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 $\textbf{eq. 1}$?

  • $\begingroup$ 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 Oct 22 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 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] $.

Hope that's the help you were looking for.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.