2
$\begingroup$

If $f$ and $g$ are uniformly continuous functions in $A$ show that $f+g$ is uniformly continuous in $A$.

Proof: because $f$ and $g$ are uniformly continuous on $A$ we can write

$$\forall\varepsilon>0,\exists\delta_1>0,\forall x,y\in A:|x-y|<\delta_1\implies|f(x)-f(y)|<\frac{\varepsilon}2$$ $$\forall\varepsilon>0,\exists\delta_2>0,\forall x,y\in A:|x-y|<\delta_2\implies|g(x)-g(y)|<\frac{\varepsilon}2$$

Then if I call $h(x)=f(x)+g(x)$ we can see that

$$|h(x)-h(y)|\le|f(x)-f(y)|+|g(x)-g(x)|$$

and if I took $\delta_0=\min\{\delta_1,\delta_2\}$ then I can finally write

$$\forall\varepsilon>0,\exists\delta_0>0,\forall x,y\in A:|x-y|<\delta_0\implies|f(x)-f(y)|+|g(x)-g(y)|<\frac{\varepsilon}2+\frac{\varepsilon}2=\varepsilon$$

My question, it is the proof correctly written? Im unsure about the use of "$\frac{\varepsilon}2$" to hold the proof. Thank you in advance.

$\endgroup$
3
$\begingroup$

The idea is correct, but since you seem to learning to write out formal proofs I will add a few critiques:

  • You say that "we can see that $|h(x)-h(y)|$...", but this feels like it's missing justification. Just a brief nod towards the triangle inequality would go a long way towards distinguishing this writeup from that of someone who was just guessing, and in a truly formal setting you would make explicit use of the triangle inequality to justify this.

  • You are right to point out that introducing $\epsilon/2$ right from the beginning is a bit odd. It's very easy to deduce that from the formal definition, but in a fully-fleshed out proof this deduction should be explicit. Starting from $\epsilon > 0$ you know that $\epsilon/2 > 0$ and thus you can apply the uniform continuity definition to $\epsilon/2$. I believe this should be spelled out (of course, if this was a journal paper, I would apply a different standard of writing).

  • There is a small stylistic disconnect between the second-last line and the conclusion. In the second-last line you write "$\delta_0 = \min\{\delta_1, \delta_2\}$" which implicitly creates a context where you have already chosen a particular $\epsilon > 0$, since otherwise $\delta_1$ and $\delta_2$ don't exist. Then in the conclusion you use "$\forall \epsilon$", which feels like you've escaped that implicit context without actually saying so.

I believe ideally you should make this context very explicit: first explain that you are choosing an $\epsilon > 0$, then deduce the existence of $\delta_1, \delta_2$ corresponding to that choice of $\epsilon$ (this would be a more traditional place at which to introduce the $\epsilon/2$), then perform the calculation showing that $|h(x)-h(y)| < \epsilon$. Having closed that figurative bracket, you are then free to say $\forall \epsilon>0$ because the preceding argument works in generality.

$\endgroup$
  • 1
    $\begingroup$ This is the kind of answer I was waiting for. Thank you very much. $\endgroup$ – Masacroso Jun 1 '16 at 1:17
1
$\begingroup$

Yes, this is right! Scaling $\varepsilon$ in these arguments by a positive constant doesn't affect the conclusions. You really just have to think about this until you convince yourself. Instead of saying "for any $\varepsilon > 0$", maybe say "for any positive number". Then, given $\varepsilon > 0$, your "positive number" can be $\varepsilon$ itself, or $\varepsilon/2$, or $3\varepsilon$, ...

and a corresponding $\delta$ exists.

$\endgroup$
  • $\begingroup$ Yes, I understand. I can write the proof with words but my question is if the formal expression in symbols is correct, this is what I was afraid of. By your answer I understand that is right in the formal expression. $\endgroup$ – Masacroso Jun 1 '16 at 0:59

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.