-4

Let $f:[0,1]\to [0,1]$ be a continuous function. Prove that $f(x) = x$ has a solution in $[0,1]$.

So, here, $f(0) = 0$ and $f(1) = 1$ and $f'(x) = 1$. Now, can we conclude that there is at least one solution and that is...

Yunxuan Zhang
  • 1,344
Jellyfish
  • 1,723
  • 5
    how do we know f(0)=0 and f(1)=1. Also f is just continuous so we could not take f'(x)... – user72012 Jul 02 '15 at 02:19
  • The problem isn't defining the function to be $f(x)=x.$ It's saying that if $f$ is any continuous function $[0,1]\to[0,1],$ then there is some $x\in[0,1]$ such that $f(x)=x.$ – Cameron Buie Jul 02 '15 at 02:25
  • Put another way, every continuous function on $[0,1]$ has at least one fixed point. – Cameron Buie Jul 02 '15 at 02:26

3 Answers3

4

Look at the function $f(x)-x$. It's nonnegative at $0$, and non-positive at $1$. So, either $f(0)-0=0$ or $f(1)-1=0$ or if none of these is the case, there must be $c$ between $0$ and $1$ such that $f(c)-c=0$, using Intermediate Value Theorem.

vgmath
  • 1,355
  • 8
  • 16
1

First of all, we do not know any of the things you suggest (consider $f(x)=x/3+1/3$). If $f$ is continuous, we know that $g(x)=f(x)-x$ is continuous. Then $g(0)\geq 0$ and $g(1)\leq 0$. By the intermediate value theorem, there exists $c\in[0,1]$ such that $g(c)=0$. Then $f(c)=c$.

Plutoro
  • 23,378
  • No, we don't even know that $g$ (or $f$) is differentiable, so we cannot consider the derivative of either. – Plutoro Jul 02 '15 at 02:25
0

I can't comment so I'm posting an answer, also in hopes that someone can correct me if I'm wrong and where.

So you ask for a proof that $f(x) = x$ has a solution in $[0;1]$, i.e. (and throwing away formalities), where $f(x)$ intercepts the $x$ axis.

But we're given that $f(0) = 0$ so there goes your trivial (as hinted by the poster) solution because, aside from the first fact, we also know for a fact thar zero is in the domain of $f$.

Isn't this enough?

Edit: I mean, if $f$ was any function... But we are given $f(x) = x$.

Giuliano
  • 25
  • 1
    We are not given $f(0) = 0$. If $f(0) = 0$ or $f(1) = 1$, we are done. Otherwise, consider the function $g(x) = f(x) - x$. If $f(0) \neq 0$, then $f(0) > 0 \Rightarrow g(0) = f(0) > 0$. If $f(1) \neq 1$, then $f(1) < 1 \Rightarrow g(1) = f(1) - 1 < 0$. Now, use the continuity of $f$ and the Intermediate Value Theorem to conclude that $g(x)$ has a zero in $[0, 1]$, which implies the desired result. – N. F. Taussig Jul 02 '15 at 08:25
  • I thought the second paragraph was kind of a second hint. That's why I became incredulous of the question itself. – Giuliano Jul 04 '15 at 00:31