1
$\begingroup$

Problem: Let $f,g:[a,b]\to [c,d]$ be continous functions. Suppose that $f$ is surjective. Prove that there exists $x\in[a,b]$ such that $f(x)=g(x)$

My solution and proof: Given $f,g:[a,b]\to[c,d]$ are both continous functions. Let $h(x)=f(x)-g(x)$. Then clearly $h$ is a continous function from $[a,b]$ into $[c,d]$, and it's the difference between two continous functions. Since $f$ is surjective, $\exists x$ such that $f(x_2)=d$. Also, $g(x_1),g(x_2)\in[c,d]$. Therefore $h(x_1)=f(x_1)-g(x_1)\leq0$ as $f(x_1)=c$ and $h(x_2)=f(x_2)-g(x_2)\geq$ as $f(x_2)=d$. Thus $h$ is a continous function having $h(x_1)\leq0$ and $h(x_2)\geq0$, then by intermediate value theorem $\exists x\in[x_1,x_2]\subseteq[a,b]$ such that $h(x)=0$ and $f(x)=g(x)$. Hence, $\exists x\in[a,b]$ such that $f(x)=g(x)$

Does my proof look correct? Anything missing?

$\endgroup$
5
  • 1
    $\begingroup$ It looks fine to me. $\endgroup$
    – José Carlos Santos
    Commented Apr 28, 2021 at 21:32
  • 2
    $\begingroup$ It's not true that $h$ is a continuous function from $[a, b]$ to $[c, d]$. In particular, the codomain isn't correct. For example, you could imagine that $c = 1$ and $d = 2$, but you're examining spots when $h(x) = 0$, and $0 \not \in [1, 2]$. $\endgroup$
    – davidlowryduda ♦
    Commented Apr 28, 2021 at 21:50
  • 1
    $\begingroup$ @davidlowryduda is correct, but this doesn't really affect your argument. Rather than figure out how small you can make the codomain of $h$, just say $h\colon[a,b]\to\mathbb R$. It's important that $\operatorname{ran}f=[c,d]$ and that $\operatorname{ran}g\subseteq[c,d]$, but $\operatorname{ran}h$ is not so important. (Well, it's important that $\operatorname{ran}h\ni0$, but the rest of it doesn't matter.) $\endgroup$
    – Toby Bartels
    Commented Apr 28, 2021 at 22:15
  • $\begingroup$ You seem to have a typo where you write ‘$\exists x$ such that $f(x_2)=d$’. You talk about $x_1$ and $x_2$, not just $x$, and use a property of $x_1$ later too. There's another typo where you have $\geq$ with nothing after it. So fix these, the codomain of $h$ (as noted above), and the missing period at the very end, and you have a good proof. $\endgroup$
    – Toby Bartels
    Commented Apr 28, 2021 at 22:21
  • $\begingroup$ One more minor error here is that you write $\exists x\in[x_1,x_2]$, but you don't know whether $x_1<x_2$ or $x_1>x_2$, so you should write $\exists x\in[\min(x_1,x_2),\max(x_1,x_2)]$ (or use words like ‘there is an $x$ between $x_1$ and $x_2$’). $\endgroup$
    – Toby Bartels
    Commented Apr 29, 2021 at 3:52

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .