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Suppose $f,g:[a,b]\to\mathbb{R}$ are continuous functions such that $f(a)\leq g(a)$ and $f(b)\geq g(b)$. Prove that $f(c)=g(c)$ for at least one $c \in [a,b]$

okay now suppose that a is less than b and $f:[a,b]\to\mathbb{R}$ is a continuous function such that the range of f contains [a,b]. Prove that f has a fixed point

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    $\begingroup$ Is this the full question? What is $c$? Are you supposed to show such a $c$ exists in $[a,b]$? $\endgroup$ – Alex R. Dec 8 '15 at 1:47
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Consider $h : [a,b] \rightarrow \mathbb{R}$ defined by $h(x) := f(x) - g(x)$. This function is continuous on $[a,b]$ and you know that $$h(a) = f(a) - g(a) \leq 0$$ and $$ h(b) = f(b) - g(b) \geq 0.$$ Now apply the Intermediate Value Theorem to $h(x)$; there exists $c \in [a,b]$ such that $h(c) = 0$. But this means that $f(c) - g(c) = 0$, i.e. $f(c) = g(c)$.

Edit :

Let's now discuss the second half. Let's suppose the trivial cases $f(a) = a$ and $f(b) = b$ do not occur.

Consider the function $g : [a,b] \rightarrow \mathbb{R}$ defined by $g(x) = x$; i.e. $g$ is the identity function on $[a,b]$. We will proceed just as in the first part by considering the continuous function $h :[a,b] \rightarrow \mathbb{R}$ defined by $h(x) := f(x) - g(x)$. We will consider two cases :

  • Suppose that $f(a) < g(a)$. Since the range of $f$ contains $[a,b]$ there exists $d \in (a,b)$ such that $f(d) = b$. Then $$f(d) = b > g(x),~~~~~~\forall x \in [a,d].$$ To sum up, $h(a) \leq 0$ and $h(d) \geq 0$. Apply the Intermediate Value Theorem to $h(x)$; there exists $c \in [a,d]$ such that $h(c) = 0$ which means that $f(c) = g(c) = c$;
  • Suppose that $f(a) > g(a)$. Since the range of $f$ contains $[a,b]$ there exists $d \in (a,b)$ such that $f(d) = a$. Then $$f(d) = a > g(x),~~~~~~\forall x \in (a,d].$$ To sum up, $h(a) \geq 0$ and $h(d) \leq 0$. Apply the Intermediate Value Theorem to $h(x)$; there exists $c \in [a,d]$ such that $h(c) = 0$ which means that $f(c) = g(c) = c$;
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  • $\begingroup$ I get it thanks! How about the other half? $\endgroup$ – user296676 Dec 8 '15 at 2:04
  • $\begingroup$ I added an answer to the other half. $\endgroup$ – M.G Jan 5 '16 at 19:17
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Let $h(x)=f(x)-g(x).$ Then $h$ is continuous and $h(a)\leq 0\leq h(b), $ so $h(c)=0$ for some $c\in [a,b].$

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  • $\begingroup$ Isn't this just rewriting the question? In other words, you didn't explain why $h(c)=0$ for some $c\in[a,b]$ $\endgroup$ – Eric S. Jan 5 '16 at 12:32
  • $\begingroup$ @Eric S .Because it's obvious . $0\in [h(a),h(b)]\subset \{h(x) :x\in [a.b]\}$ because $h$ is continuous. Continuous functions have the intermediate value property. $\endgroup$ – DanielWainfleet Jan 5 '16 at 20:43
  • $\begingroup$ I know it's obvious, just like the original question is obvious. But proving such an obvious statement should indeed refer to the intermediate value theory $\endgroup$ – Eric S. Jan 6 '16 at 10:07

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