Let $f:[a,b]\rightarrow \mathbb{R}$ bounded and

$\omega(f,r)=\sup\{|f(x)-f(y)| \colon x,y \in [a,b], \ |x-y|<r\}$

(called modulus of continuity of $f$ EDIT note: the original question is in spanish, I'm not sure if the "modulus of continuity" is a proper translation from "módulo de continuidad")

Show that the following properties of $\omega(f,r)$ are satisfied:

  1. if $0<r_1 < r_2$ then $\omega(f,r_1)<\omega(f,r_2)$
  2. $\omega(f,r_1+ r_2) \leq \omega(f,r_1) + \omega(f,r_2)$
  3. $f$ is uniformly continuous if and only if $\lim_{r \searrow 0}\omega(f,r)=0$
  4. If $\lambda>0$ then $\omega(f,\lambda r)< (1+\lambda)\omega(f,r)$

So far I have proved only the first one:

  1. Let $\gamma(r)=\{|f(x)-f(y)|\colon x,y \in [a,b], \ |x-y|<r\}$

    Then $\gamma(r_1)\subseteq \gamma(r_2)$

    Then $\sup\gamma(r_1) \leq \sup \gamma(r_2)$ (*)

    $\therefore \omega(f,r_1)\leq \omega(f,r_2)$

(*) Let $A,B \subseteq \mathbb{R}$ bounded above, such that $A \subseteq B$, we know that $\sup B$ is an upper bound of $A$ and by definition, $\sup A$ is the smallest upper bound of $A$ $\therefore \sup A \leq \sup B$.

  1. For the second I am trying to use a triangle inequality but I'm having a hard time incorporating it inside $\omega(f,r)$ if that makes any sense. I'm not exactly sure how I should start.

  2. For this one I tried using the definition of uniform continuity using sequences, but I'm not sure I'm writing it properly:

    $\forall \varepsilon >0$ $\exists \ (X_n)_{n=1}^\infty ,(Y_n)_{n=1}^\infty $ sequences in $[a,b]$, such that $|X_n-Y_n|<\frac{1}{n} \Rightarrow$ $|f(X_n)-f(Y_n)|<\varepsilon$

  3. I don't know how to start with this one

Any corrections, comments, suggestions are highly appreciated.

  • 2
    $\begingroup$ #2: Notice that if $|x-y|<r_1+r_2$, then there exists some $z$ such that $|x-z|<r_1$ and $|y-z|<r_2$. An inequality can be used to get rid of mention of the $z$'s. #3: That is not the correct definition for uniform convergence. #4: Are you sure the inequality is strict? $\endgroup$ – minimalrho Sep 12 '14 at 3:28
  • 1
    $\begingroup$ Thanks for the suggestions, yes the inequality is strict. $\endgroup$ – Adriana LE Sep 12 '14 at 3:47
  1. you are right since $|x-y|<r_1$ implies $|x-y|<r_2\,$
    but I don't see the strict inequality;

  2. following @minimalrho one obtains $$|f(x)-f(y)|\le|f(x)-f(z)|+|f(z)-f(y)|\le\omega(r_1)+\omega(r_2)$$

  3. if: the existence of $\delta>0$ such that $ \omega(r)<\varepsilon$ for all $r<\delta$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y$ such that $|x-y|<\delta$, the definition of uniform continuity;

    only if: suppose $\omega(r)$ does not tend toward $0$ as $r \to 0\,$. Since by 1. $\omega$ is monotonic, there exists $\varepsilon>0$ such that $\omega(r)\ge \varepsilon$ for all $r>0$. But the uniform continuity implies the existence of $\delta>0$ such that $|f(x)-f(y)|<\varepsilon/2$ for all $x,y$ such that $|x-y|<\delta$, contradicting the previous statement;

  4. I agree with @minimalrho about the strict inequality (perhaps there is a typo);
    from 3., if $n$ is a natural number, one obtains by induction $\omega(nr) \le n \omega(r)$;
    taking the nonnegative integer $n$ for which $n \le\lambda<n+1$, one sees that $$\omega(\lambda r)\le\omega((n+1) r)\le (n+1)\omega(r)\le (\lambda+1)\omega(r)$$

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