Iv'e got some questions involving uniform continuity of functions and its properties. I would like that someone will check my work and maybe help with correcting flaws.

Let $A \subset R$ and let $f,g: \ A \rightarrow \mathbb{R} $ be uniformly continuous in A.

Prove or disprove:

  • f+g is uniformly continuous in A:

    By definitions:

    • $\forall \epsilon'>0 \ \exists \delta_1>0: \ |x-y|<\delta_1 \Rightarrow |f(x)-f(y)|<\epsilon'$

    • $\forall \epsilon'>0 \ \exists \delta_2>0: \ |x-y|<\delta_2 \Rightarrow |g(x)-g(y)|<\epsilon'$

Now, let $h(x)=f(x)+g(x)$ and $\delta=min(\delta_1,\delta_2)$.

By the triangle inequality we may write the following: $|[f(x)-f(y)]+[g(x)-g(y)]| \leq |f(x)-f(y)|+|g(x)-g(y)|<\epsilon'+\epsilon'=2\epsilon'$.

Let $\epsilon'=\frac{\epsilon}{2}$ we get eventually: $\forall \epsilon>0 \ \exists \delta>0: \ |x-y|<\delta \Rightarrow |h(x)-h(y)|<2\epsilon'=\epsilon$.

  • $f\cdot{g}$ is uniformly continuous in A - This is false as we may take $f(x)=x, \ g(x)=x$.
  • f,g are bounded. $f\cdot{g}$ is uniformly continuous in A - I have seen proofs here and here.
  • Given $A=\mathbb{R}$. $f \circ g$ is uniformly continuous in $\mathbb{R}$ - I have no clue - no matter what I tried to do, I get to dead end.

Please help, thanks!

  • $\begingroup$ Note that $\delta={\epsilon}$. $\endgroup$ – Mhenni Benghorbal Dec 25 '13 at 15:28
  • $\begingroup$ How does it matter? where should I include it in my solution? Have I written the solution in a correct proper way? $\endgroup$ – Galc127 Dec 25 '13 at 15:33
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    $\begingroup$ Your Proof is ok. Just I wanted to point out the relation between $\epsilon$ and $\delta$. See for example here. $\endgroup$ – Mhenni Benghorbal Dec 25 '13 at 15:39
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    $\begingroup$ Mhenni, thanks, now it is understood. David, I'm familiar with the definitions and tried to start from definitions and elaborate, but no success. May you continue? $\endgroup$ – Galc127 Dec 25 '13 at 15:45
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    $\begingroup$ So, continuing with what I wrote, if $|x-y|<\delta_2$, it follows that $|f(g(x))-f(g(y))|<\epsilon$. As $\epsilon$ was arbitrary, it follows that $f\circ g$ is uniformly continuous. $\endgroup$ – David Mitra Dec 25 '13 at 15:49

Seems fine to me.

For your last question about the composition of $f$ and $g$. It might be wise to make a small sketch. $$x \overset{\delta_g}{\longrightarrow}g(x) \overset{\epsilon_g=\delta_f}{\longrightarrow}f(g(x))$$ If I recall it correctly.

  • $\begingroup$ It supposed to be $f \circ g=f(g(x))$, but your hint is pretty good and helpful. Thank you. $\endgroup$ – Galc127 Dec 25 '13 at 15:47
  • $\begingroup$ You are right, fixed it. $\endgroup$ – Nigel Overmars Dec 25 '13 at 15:49

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