# Are continuous functions with compact support bounded?

While studying measure theory I came across the following fact: $\mathcal{K}(X) \subset C_b(X)$ (meaning the continuous functions with compact support are a subset of the bounded continuous functions). This seems somehow odd to me; I've tried to prove it but did not succeed. Could someone help me out here?

Thanks!

• Let $f \in \mathcal{K}(X)$, and $S$ the support of $f$. What do you know about $f(S)$? And what about $f(X\setminus S)$? – Daniel Fischer Jun 30 '15 at 15:41
• Your question is in fact Weierstrass's theorem. Have you studied it or not? This is in order for the answerers to know at what level to "calibrate" their arguments. – Alex M. Jun 30 '15 at 15:55
• In any topology, the image of a compact set is compact (under a continuous map). – MartianInvader Jun 30 '15 at 19:29

We have $$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)),$$ which is compact in $\mathbb{R}$ (since $\mathrm{supp}(f)$ is compact and $f$ is continuous), hence bounded.

• And why is that compact? It seems to me that your proof is circular, assuming exactly what the OP wants to prove. – Alex M. Jun 30 '15 at 15:51
• @AlexM. Because the image of a compact set by a continuous function is compact and the union of two compact sets is compact. – Spenser Jun 30 '15 at 15:52
• Assuming that the OP knows that the image of a compact through a continuous function is again compact, then yes. But since the OP doesn't even know Weierstrass's theorem (which is simpler), I suspect that he asks for a much more basic proof, one close to first principles. – Alex M. Jun 30 '15 at 15:54
• Technically, isn't it true that $f(X) = \{0\} \cup f(\text{supp}(f))$? Then we can see that it is compact directly by compactness of $\{0\}$ and $f(\text{supp}(f))$ (and the fact that the finite union of compact sets is compact). – Satana Mar 24 at 1:08
• I take my above comment back, since $\{x\in X: f(x) =0\}$ need not be non-empty. Nevertheless, if $\{x\in X: f(x) =0\}\ne \emptyset$, then $f(X) = \{0\} \cup f(\text{supp}(f))$. On the other hand, if $\{x\in X: f(x) =0\}= \emptyset$, then $f(X) = f(\text{supp}(f))$. In either case, $f(X)$ is compact. – Satana Mar 24 at 1:32

If you do know the Weierstraß theorem, then you can prove it like that:

Let $f \in \mathcal{K}(X)$ and denote by $K$ the support of $f$. Then $f|_{K^c}=0$ by the very definition of the support. Moreover, by the Weierstraß theorem, $f|_K$ is bounded. Combining both facts, proves that $f$ is bounded.

If you do not know the Weierstraß theorem, then have e.g. a look at this answer (note that the proof does not only work for a compact interval $[a,b]$, but for any compact set $K$).

• This is a circular "proof", because the OP wants precisely the proof of Weierstrass's theorem. Otherwise what meaning would the question have? – Alex M. Jun 30 '15 at 15:50
• @AlexM. Well, I think that the OP is simply not aware of the fact that Weierstraß theorem can be applied. (Did you downvote both answers?) – saz Jun 30 '15 at 15:53
• Yes, the downvotes are mine. See the OP's comment under Spenser's answer. – Alex M. Jun 30 '15 at 16:01
• @saz $-1+1=0$ ;-) – Spenser Jun 30 '15 at 16:23
• @AlexM. I see, but it's hardly our fault. If the OP doesn't given any details what he has tried or what kind of (mathematical) background he has, then it's pretty hard to answer the question properly, right? – saz Jun 30 '15 at 16:47