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\centerline{\bf Math 313 , Introduction to Analysis }
\centerline{uniform continuity - worked example 1}

\bigskip
 
   {\bf Example:} Show that the function $f(x) = x^2$ on the closed interval $[1,3]$ is uniformly continuous.
   
   {\bf Solution:} Given $\epsilon >0$ we want to find $\delta > 0$ that works for every point in the interval $[1,3]$.  
   
   This means that for all $1 \leq c \leq 3$ and any $| h | < \delta$, we have $|f(c+h) - f(c) | < \epsilon$. 
   
   \underline{Step 1} - use the definition of $f(x)$ to reduce what we are trying to obtain to a simpler form
  $$|f(c+h) - f(c) | = |(c+ h)^2 - c^2| = |(c^2 + 2ch + h^2) - c^2|  = |2ch + h^2| \leq |2ch| + |h^2|$$
 So it is enough  to find $\delta > 0$ small enough so that the two terms on the right side above    satisfy
   $ |2ch| < \epsilon/2$ and $h^2 < \epsilon/2 $
   for every $1 \leq c \leq 3$.
   
 \underline{Step 2} -  find a condition that makes   $ |2ch| < \epsilon/2$.  Use that    $c \leq 3$, so $|2ch| \leq |2 \cdot 3 \cdot h| = 6 |h|$. 
  
So if $6 |h| < \epsilon/2$, or $|h| < \epsilon/12$,  then    $ |2ch| < \epsilon/2$ is satisfied. 
   
 \underline{Step 3} -   find a condition on $\delta$ which makes  $h^2 < \epsilon/2$.  Take   square roots of both sides  to see that we need  $|h| < \sqrt{\epsilon/2}  = \sqrt{\epsilon}/\sqrt{2}$. 
There is no $c$ in this equation, so   assume  $\delta   \leq \sqrt{\epsilon}/\sqrt{2}$ and this works for any value of $c$.
 
      
   \underline{Step 4} -   we need both upper bounds on $\delta$ to be true, so  take   $\delta = \min \{ \sqrt{\epsilon}/\sqrt{2} , \epsilon/12\}$.   That'll do it.
    
   Since  $\delta \leq \sqrt{\epsilon}/\sqrt{2}$ then $|h| < \delta$ implies   $h^2 < \epsilon/2$.  
   
   Since  $\delta \leq \epsilon/12$ then $|h| < \delta$ implies  $|6h| < \epsilon/2$ so that $|2ch| < \epsilon/2$ for all $1 \leq c \leq 3$. 
   
   Combine these two, so $|h| < \delta$ implies 
    $$|f(c+h) - f(c) | = |(c+ h)^2 - c^2| =  |2ch + h^2| \leq |2ch| + |h^2| < \epsilon/2 + \epsilon/2 = \epsilon$$
 

    
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