SOLUTION TO PROBLEM 5:

Let $g(x)=f(x)-f(x+\frac{1}{n})$. Then it is sufficient to find $x\in [0, 1-\frac{1}{n}]$ such that g(x)=0. Notice also that since f is continuous on [0,1], the mapping $x\mapsto f(x+\frac{1}{n})$ is continuous on $[-\frac{1}{n},1-\frac{1}{n}]$, and therefore the functionction g is continuous on the interval $[0,1-\frac{1}{n}]$. Now note that

$$\begin{array}{rcl} g(0)&=& f(0)-f(\frac{1}{n}),\\ &&\\ g(\f... ...\ &&\\ g(\frac{n-1}{n})&=&f(\frac{n-1}{n})-f(1). \end{array}$$

When all of the equations are added together, the right side will telescope to get

$$g(0)+g\Big(\frac{1}{n}\Big)+g\Big(\frac{2}{n}\Big)+\ldots+g\Big(\frac{n- 2}{n}\Big)+g\Big(\frac{n-1}{n}\Big)= f(0)-f(1),$$

and thus

$(\circledast)$

         $$g(0)+g\Big(\frac{1}{n}\Big)+ g\Big(\frac{2}{n}\Big)+\ldots+g\Big(\frac{n-2}{n}\Big)+g\Big(\frac{n-1}{n} \Big)=0$$.

Now, if any $g\Big(\frac{i}{n}\Big)$ is zero, we are finished (this is guaranteed to be the case for n=1 since then g(0)=f(0)-f(1)=0). Notice that it cannot be the case that all $g\Big(\frac{i}{n}\Big)$ are positive or all are negative since then $(\circledast)$ would not be satisfied. Plainly, this implies that there are two adjacent $g\Big(\frac{i}{n}\Big)$ which differ in sign. By the Intermediate Value Theorem, g(x) must have root between those two values.
Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Sep 29 13:29:48 CDT 2000