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Can you tell me if the following proof is correct?

Claim:

If $f$ is a continuous and compactly supported function from a metric space $X$ into $\mathbb{R}$ then $f$ is uniformly continuous.

Proof:

The proof is in two parts. First we want to show that $f$ is uniformly continuous on $K := \operatorname{supp}{f}$:

Let $\varepsilon > 0$. Because $f$ is continuous we have that for each $x$ in $K$ there is a $\delta_x$ such that for all $y$ with $d(x,y) < 2 \delta_x$ we have $|f(x) - f(y)| < \varepsilon$ and because $\{ B(x, \frac{\delta_x}{2}) \}_{x \in K}$ is an open cover of $K$ there is a finite subcover which we denote $\{ B(x_i, \frac{\delta_i}{2}) \}_{i=1}^n$. Define $\delta := \min_i \frac{\delta_i}{2}$ and let $x$ and $y$ be any two points in $K$ with $d(x,y) < \delta$. $\{ B(x_i, \frac{\delta_i}{2}) \}_{i=1}^n$ is a cover so there exists an $i$ such that $x$ is in $B(x_i, \frac{\delta_i}{2})$ which means that $d(x,x_i) < \frac{\delta_i}{2}$. Then $d(x_i ,y) \leq d(x_i ,x) + d(x,y) < \frac{\delta_i}{2} + \delta \leq \delta_i$ hence $y$ is also in $B(x_i, \delta_i)$.

Since $d(x_i,y) < \delta_i$ and $d(x, x_i) < \delta_i$ we have $|f(x) - f(y)| \leq |f(x) - f(x_i)| + |f(x_i) - f(y)| < 2 \varepsilon$.

Next we want to show that if $f$ is uniformly continuous on $K$ then it is uniformly continuous on all of $X$:

Let $\varepsilon > 0$. For any two points $x$ and $y$ we're done if either both are in $K$ or both are outside $K$ so let $x \in X \setminus K$ and $y \in K$ with $d(x,y) < \delta$. Then there is an $i$ such that $y$ is in $B(x_i, \frac{\delta_i}{2})$. Then $d(x,x_i) \leq d(x,y) + d(y,x_i) < \delta_i$ and hence $|f(x) - f(y)| \leq |f(x) - f(x_i)| + |f(x_i) - f(y)| < 2 \varepsilon$.

Is it necessary to prove this in two parts or is the second part "obvious" and should be left away?

Thanks for your help.

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5  
I'm more wondering whether the first part is unnecessarily detailed. Don't you have a canned theorem that a continuous function with compact domain is uniformly continuous which you can just apply to $f|_K$? –  Henning Makholm Jan 2 '12 at 0:19
3  
In the sentence "So $x,y \in B(x_i, \delta_i)$ which means that $d(x,y) < 2 \delta_i$ and therefore $|f(x) - f(y)| < \varepsilon$." where does this come from? You know that $|f(x) - f(y)| \lt |f(x)-f(x_i)| + |f(x_i)-f(y)| \lt 2\varepsilon$, which is good enough, but I fail to see how you get $\lt \varepsilon$. Is this a typo, or am I missing something? –  t.b. Jan 2 '12 at 0:24
    
@t.b. That was a "thinko", not a typo. Yes, of course it's less than $2 \varepsilon$. Thank you! –  Matt N. Jan 2 '12 at 9:29
    
@Matt You have not quite corrected the "thinko" t.b. pointed out. You say that since $d(x,y) < \delta_i$ you know that $|f(x)-f(y)|$ is small by assumption. In fact you have not made any such assumption. The assumption you made is that $|f(x)-f(y)|$ is small if $d(x,y)<\delta_x$. –  Johan Jan 2 '12 at 10:24
    
@Johan For a continuous function $f$ I have that for an $\varepsilon$ I can find a $\delta$ such that $f(B(x, \delta)) \subset B(f(x), \varepsilon)$. Above I have chosen $\delta_i$ such that $f(B(x_i, \delta_i)) \subset B(f(x_i), \varepsilon)$. I also have $x,y \in B(x_i,\delta_i)$ so $f(x), f(y) \in f(B(x_i,\delta_i)) \subset B(f(x_i), \varepsilon)$ and therefore $|f(x) - f(y)| < 2 \varepsilon$. Can you point out my thinko more explicitly please? I still don't see it, thank you. –  Matt N. Jan 2 '12 at 14:04

1 Answer 1

up vote 3 down vote accepted

That looks good except for the correction that t.b. pointed out. In the spirit of Henning Makholm's comment, here is a "canned theorem" approach.

A continuous function on a compact metric space is uniformly continuous, so $f|_K$ is uniformly continuous. Let $\varepsilon>0$ be given. Then $K_\varepsilon:=\{x:|f(x)|\geq \varepsilon\}$ is a closed subset of $K$, hence compact, and $\{x:f(x)=0\}$ is a closed set disjoint from $K_\varepsilon$, so there is a positive distance $\delta_1$ between $K_\varepsilon$ and $\{x:f(x)=0\}$. Let $\delta_2$ be such that if $x$ and $y$ are in $K$ and $d(x,y)<\delta_2$, then $|f(x)-f(y)|<\varepsilon$. Let $\delta=\min\{\delta_1,\delta_2\}$.

If $x$ and $y$ are in $X$ and $d(x,y)<\delta$, then:

  • $x$ and $y$ are both in $K$, and since $d(x,y)<\delta_2$, we have $|f(x)-f(y)|<\varepsilon$; or
  • one of $x$ or $y$ is not in $K$. WLOG suppose $x$ is not in $K$. Then $f(x)=0$, and since $d(x,y)<\delta_1$, $y$ is not in $K_\varepsilon$, meaning $|f(x)-f(y)|=|f(y)|<\varepsilon$. $\square$
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