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How do you show that there is a solution to a function that may not be continuous using a similar concept to IVT?

Juniven Acapulco
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Jay
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4 Answers4

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Let $A = \{x\in [0,1]\mid f(x)\geq x\}$. This set is bounded (by $1$), and non-empty (since $0 \in A$), and thus has a least upper bound, say $a = \sup A$. I now disprove the following two statements by contradiction:

  1. $f(a) > a$. If this were true, then since $f$ is increasing we would get $f(f(a))\geq f(a)$. This means $f(a) \in A$, which contradicts $a$ being an upper bound of $A$.
  2. $f(a) < a$. If this were true, and $a$ is an upper bound for $A$, then for any $x \in (f(a), a)$, we have $f(a)<x$ and $x<a$, which since $f$ is increasing implies $f(x)\leq f(a)$. Together this makes $f(x)<x$. This implies that $f(a)$ is also an upper bound for $A$ (in other words, $A$ has no elements in the interval $(f(a),a]$), which contradicts $a$ being the least upper bound.

Thus we are forced to conclude that $f(a) = a$.

Arthur
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  • Would you mind sharing some of your lines of thought? :) –  Jan 25 '17 at 14:43
  • @OpenBall I rephrased it a bit. Is there still something you think I'm missing? – Arthur Jan 25 '17 at 15:15
  • I wasn't saying that your answer is missing something, but that you introduced $A$ quite out of nowhere. Of course that's completely fine and has the advantage of making your answer elegant, but it would be great if you might share how you thought it out :) –  Jan 25 '17 at 15:19
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    @OpenBall I named it $A$ in order to be able to reference it later. The set itself makes sense to consider because $f$ starts out above $x$ and ends up below it. That, coupled with a limitation on how quickly $f$ can move across the line $x$ in that direction ($f$ can just jump upwards, but downwards it has to wait for $x$ to catch up and go past) meant that studying such a crossing felt natural. The last place it happens is $a$, and I think that that's the easiest one to formulate specifically, seeing as $\sup$ is so handy and all :-) – Arthur Jan 25 '17 at 15:25
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If $f(0) = 0$ or $f(1) = 1$, we are done, so suppose this is not the case.

Then, $f(0) > 0$ and $f(1) < 1$.

Now, consider the function $g(x) = x - f(x)$. Here, $g(0) < 0$ and $g(1) > 0$.

Then, either $g(y) = 0$ for some value of $y \in (0,1)$ or there exists $y \in (0,1)$ such that

$\forall \epsilon > 0, \exists x$ such that $y-\epsilon < x < y$ and $g(x) < 0$ while $g(y) > 0$.

Let $g(y) = \delta > 0$ and pick $\epsilon < \delta$. Then if we pick $x$ with $y-\epsilon < x < y$ and $g(x) < 0$, we have

$f(x) > x > y-\epsilon > y-\delta = f(y)$ which contradicts the fact that $f$ is increasing

hexomino
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Suppose that it isn't true, so there's no $x\in[0,1]$ for which $f(x)=x$. Define $A=\{x\in[0,1]\colon f(x)>x\}$ and $B=\{x\in[0,1]\colon f(x)<x\}$. Both these sets aren't empty ($0\in A$, $1\in B$). Consider $z=\inf B$.

  1. If $z\in A$ then $f(z)>z$. But since $z=\inf B$ then there exists $x\in B$ such that $z<x<z+(f(z)-z)=f(z)$ for which $f(x)<x<f(z)$ which contradicts the fact that $f(x)$ is increasing.

  2. If $z\in B$ then $z>0$, and $f(z)<z$. Similarly, there exists $x\in A$ such that $f(z)<x<z$, for which $f(x)>x>f(z)$ (again, contradiction).

Sergei Golovan
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Let $g(x)=f(x)-x$. Then $g(0)=f(0)>0$ (assuming $f(0)\ne 0$ and $f$ is increasing) and $g(1)=f(1)-1<0$ (assuming $f(1)\ne 1$ and $f$ is bounded above by $1$) which suggests that $g(x)=f(x)-x=0$ has at least one root in the interval $(0,1)$$\implies f(x)=x$ is satisfied at least once in $(0,1)$.

Nitin Uniyal
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