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The main idea here is to apply the fixed point theorem to $g(x)=f(x)+x$, in order to show that f has a zero in $I$. If $g$ has a fixed point (i.e. $g(x_0)=x_0$), then $f(x_0)=0$. I just don't see how to show that $d(g(x),g(y))=d(f(x)+x,f(y)+y) \leq kd(x,y)$ for all $x,y \in I$, and with $0<k<1$, which is necessary to apply the fixed point theorem to $g$.

EDIT:

This is an exercise designed to help me learn how to use the following fixed point theorem:

If $f$ is a mapping on a complete metric space $S$ into $S$ such that $d(f(x),f(y)) \leq kd(x,y)$ for all $x,y \in S$ with $0<k<1$, then the mapping has a unique fixed point.

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    $\begingroup$ Isn't the intermediate value theorem sufficient? $f(0)=1/4>0$ and $f(1)=-1/40<0$. $\endgroup$ – Daniel Mar 23 '15 at 22:26
  • $\begingroup$ $f(0)=\frac14$ and $f(1)=-\frac{1}{40}$. $f$ is continuous and so, there must be a root on $(0,1)$. $\endgroup$ – Mark Viola Mar 23 '15 at 22:27
  • $\begingroup$ @DanielEscudero I edited the title. This is a problem to help me learn about the fixed point theorem. $\endgroup$ – ztforster Mar 23 '15 at 22:28
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    $\begingroup$ @ThomasAndrews Thanks, I didn't realize that. I've clarified with an edit. $\endgroup$ – ztforster Mar 23 '15 at 22:36
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    $\begingroup$ More apologies. It's fixed $\endgroup$ – ztforster Mar 23 '15 at 22:39
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By the MVT, $g(x)-g(y) =g'(\xi) (x-y)$ for $0<\xi<1$. Thus,

$$\begin{align} |g(x)-g(y)| & =|g'(\xi)| |(x-y)| \\ & = \left(\frac34 -\frac18 \xi^4\right) |x-y| \\ & \le \frac34 |x-y| \\ & < (1) |x-y| \end{align}$$

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    $\begingroup$ This works just fine for me, you were just beaten for time :) $\endgroup$ – ztforster Mar 23 '15 at 23:04
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    $\begingroup$ My pleasure. I got a tighter bound for you! $\endgroup$ – Mark Viola Mar 23 '15 at 23:05
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$$\begin{align}|g(x)-g(y)|&=\left|\frac{3}{4}(x-y)-\frac{1}{40}(x^5-y^5)\right|\\ &\leq\frac{3}{4}\left|x-y\right|+\frac{1}{40}\left|x^5-y^5\right|\\ &=\frac{3}{4}\left|x-y\right|+\frac{1}{40}\left|x-y\right|(x^4+x^3y+x^2y^2+xy^3+y^4)\\ &\leq\frac{3}{4}\left|x-y\right|+\frac{5}{40}\left|x-y\right|\\ &=\frac{35}{40}\left|x-y\right|\end{align}$$

You can take $k=35/40 = 7/8<1$.

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  • $\begingroup$ Really valuable to learn the latex \begin{align}...\end{align} for this sort of expression. $\endgroup$ – Thomas Andrews Mar 23 '15 at 22:39
  • $\begingroup$ Do you mean that way? Mmm I don't see it better... Should a \displaystyle fix it? $\endgroup$ – Daniel Mar 23 '15 at 22:40
  • $\begingroup$ I've just done it for you. Hope you don't mind. $\endgroup$ – Thomas Andrews Mar 23 '15 at 22:41
  • $\begingroup$ Of course I don't! I was trying it, but you were faster. Thanks a lot for the advice! I really appreciate it! $\endgroup$ – Daniel Mar 23 '15 at 22:42
  • $\begingroup$ You need one more fact for this to be complete - you need $0\leq g(x)\leq 1$ for $0\leq x\leq 1$. $\endgroup$ – Thomas Andrews Mar 23 '15 at 22:58

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