1. ## Uniform continuity question

A few of my friends had enrolled for an Analysis-I course and had recently written their final exams. One of the questions they were asked to prove was(I am writing it from memory, so forgive me if its meaningless):

If f is uniformly continuous over $\mathbb{R}$, then prove that there exists constants a and b such that $|f(x)| \leq a|x|+b, \forall x \in \mathbb{R}$

I had not enrolled for the course, but I know the definition of uniform continuity and a couple of simple theorems about it. This problem looked too harsh on uniformly continuous functions and I thought it was a misprint ......But it was not

Any ideas on how to solve it?
Do I need anything other than the definition? if yes then please tell me the proof. Else give me a hint.

Thank you,
Srikanth

2. Definition of uniform continuity: $\forall\epsilon>0\;\exists\delta>0$ such that $|x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<\varepsilon$. Apply this with $\varepsilon=1$ to get a $\delta>0$ with $|x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<1$. In particular, with y=0, the triangle inequality gives $|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|$.

Suppose in addition that $|x|>\delta/2$. Then $\delta/2<|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|< \tfrac2{\delta}|x| + |f(0)|$.

Next (assuming from now on that $\delta/2<|x|<\delta$), $|2x-x|<\delta$ and so $|f(2x)-f(x)|<1$. By the triangle inequality again, $|f(2x)|<2+|f(0)|<\tfrac2{\delta}|2x| + |f(0)|$.

That should suggest an inductive argument: $|f(nx)|<\tfrac2{\delta}|nx| + |f(0)|$.

But multiples of the interval $[-\delta/2,\delta]$ cover all the positive reals except for the interval $[0,\delta/2]$. So take $a=\tfrac2{\delta}$ and $b=1+|f(0)|$, and you'll have $|f(x)|\leqslant a|x|+b$ for all real x.

3. Hi !
Originally Posted by Opalg
$|f(x)-f(y)|<1$. In particular, with y=0, the triangle inequality gives $|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|$.
That's the part that stuck me
How can the triangle inequality help ?
Doesn't it say $||x|-|y|| \leq \dots$ ?
So since it's $\leq$, we cannot know if the "right parts" will still be < to 1 ?
Can you enlighten me pleaaaase ?

4. Originally Posted by Moo
How can the triangle inequality help ?
Doesn't it say $||x|-|y|| \leq \dots$ ?
So since it's $\leq$, we cannot know if the "right parts" will still be < to 1 ?
If $|a-b|<1$ then $|a| = |(a-b)+b|\leqslant|a-b|+|b|<1+|b|$.

5. Originally Posted by Opalg
Definition of uniform continuity: $\forall\epsilon>0\;\exists\delta>0$ such that $|x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<\varepsilon$. Apply this with $\varepsilon=1$ to get a $\delta>0$ with $|x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<1$. In particular, with y=0, the triangle inequality gives $|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|$.

Suppose in addition that $|x|>\delta/2$. Then $\delta/2<|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|< \tfrac2{\delta}|x| + |f(0)|$.

Next (assuming from now on that $\delta/2<|x|<\delta$), $|2x-x|<\delta$ and so $|f(2x)-f(x)|<1$. By the triangle inequality again, $|f(2x)|<2+|f(0)|<\tfrac2{\delta}|2x| + |f(0)|$.

That should suggest an inductive argument: $|f(nx)|<\tfrac2{\delta}|nx| + |f(0)|$.

But multiples of the interval $[-\delta/2,\delta]$ cover all the positive reals except for the interval $[0,\delta/2]$. So take $a=\tfrac2{\delta}$ and $b=1+|f(0)|$, and you'll have $|f(x)|\leqslant a|x|+b$ for all real x.

Thanks Opalg. The proof is wonderful since does not use any "theorems". The tricks you have used are fantastic...