What can I say about y-uniform continuity

[evinda]

Hello! (Wasntme)

I am looking at the following exercise:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous function with the following property:
$\forall \epsilon >0 \exists M=M( \epsilon)>0 \text{ such that if } |x| \geq M \text{ then } |f(x)|< \epsilon$.
Show that $ f$ is uniformly continuous.

That's what I have tried so far:

Let $\epsilon'>0$

• $x,y \in [-M,M] :$
  
  $f$ is continuous on $[-M,M]$,so $f$ is also uniformly continuous on this interval.So,$\exists \delta=\delta(\epsilon')>0 \text{ such that } \forall x,y \in [-M,M] \text{ with } |x-y|<\delta , \text{ we have } |f(x)-f(y)|< \epsilon'$
  
• $x,y \geq M:$
  $\exists \delta=\delta(\epsilon')>0 \text{ such that } \forall x,y \in [M,+\infty) \text{ with } |x-y|<\delta \text{ we have } |f(x)-f(y)| \leq |f(x)|+|f(y)|<2 \epsilon'$
  
• $x,y \leq -M: $
  $\exists \delta=\delta(\epsilon')>0 \text{ such that } \forall x,y \in (-\infty,-M] \text{ with } |x-y|<\delta \text{ we have } |f(x)-f(y)| \leq |f(x)|+|f(y)|<2 \epsilon'$

Is it right so far,or have I done something wrong? (Thinking)

Also,then I have to check the cases $x<-M<y$ and $x<M<y$

For the case $x<-M<y$,that's what I have done:

$x<-M \Rightarrow |f(x)|< \epsilon' $

But,what can I say about $y$,in order to conclude that $|f(x)-f(y)|< \epsilon'$ ? (Thinking) (Thinking)

 

[Opalg]

Hello! (Wasntme)

I am looking at the following exercise:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a continuous function with the following property:
$\forall \epsilon >0 \exists M=M( \epsilon)>0 \text{ such that if } |x| \geq M \text{ then } |f(x)|< \epsilon$.
Show that $ f$ is uniformly continuous.

That's what I have tried so far:

Let $\epsilon'>0$

• $x,y \in [-M,M] :$
  
  $f$ is continuous on $[-M,M]$,so $f$ is also uniformly continuous on this interval.So,$\exists \delta=\delta(\epsilon')>0 \text{ such that } \forall x,y \in [-M,M] \text{ with } |x-y|<\delta , \text{ we have } |f(x)-f(y)|< \epsilon'$
• $x,y \geq M:$
  $\exists \delta=\delta(\epsilon')>0 \text{ such that } \forall x,y \in [M,+\infty) \text{ with } |x-y|<\delta \text{ we have } |f(x)-f(y)| \leq |f(x)|+|f(y)|<2 \epsilon'$
• $x,y \leq -M: $
  $\exists \delta=\delta(\epsilon')>0 \text{ such that } \forall x,y \in (-\infty,-M] \text{ with } |x-y|<\delta \text{ we have } |f(x)-f(y)| \leq |f(x)|+|f(y)|<2 \epsilon'$

Is it right so far,or have I done something wrong? (Thinking)

Also,then I have to check the cases $x<-M<y$ and $x<M<y$

For the case $x<-M<y$,that's what I have done:

$x<-M \Rightarrow |f(x)|< \epsilon' $

But,what can I say about $y$,in order to conclude that $|f(x)-f(y)|< \epsilon'$ ? (Thinking) (Thinking)

This problem is similar to the one in http://mathhelpboards.com/analysis-50/bounded-derivative-uniform-continuity-10997.html, and you have already provided most of the solution. The remaining part of the problem is to deal with the case where one of the points $x,y$ lies inside the interval $[-M,M]$ and the other one lies outside it. My suggestion is to use the fact that $f$ is (uniformly) continuous on the interval $[-(M+1),M+1]$, and then to choose your $\delta$ to be less than $1$. Then both of $x$ and $y$ will either be inside the "expanded" interval $[-(M+1),M+1]$ or outside the original interval $[-M,M]$. So you will no longer have the problem of one point being inside the interval and the other one outside it.