Applying the IVT on closed intervals

Question

I'm not exactly sure how the intermediate value theorem can be applied to the following question especially with the defined function $g(x)$ which is provided below.

Using the intermediate value theorem with $g(x)=f(x)-x$ and suppose $f:[0,b] \to [0,b]$ is continuous. Prove there is at least one $c \in [0,b]$ such that $f(c)=c$.

Thanks

Answer 1

If $f(0)=0$, then $g(0)=f(0)-0=0$, so $x=0$ is the root of $g(x)=0$,. Next if $f(b)=b$ then $g(b)=b-b=0$, so $x=b$ is a root of $g(x)=0$. In other cases $g(0)=f(0)-0 >0, g(b)=f(b)-b <0$, so by IVT it follows that $g(x)=0$ has a root in (0,b)$.

Answer 2

Let's think through this backwards.

• "there is at least one $c \in [0,b]$ such that $f(c)=c$." is certainly the correct form of the conclusion of the IVT. So to obtain this, we need to satisfy the IVT's hypotheses.
• If there is a $c$ such that $f(c) = c$, then $f(c) - c = 0$. So, we want to know if $f(x) - x$ has a root, say at $x = c$. Conveniently, we have already been told to study this function, $g(x) = f(x) - x$.
• The hypotheses of the IVT require that $g$ is continuous and that $g$ takes at least one positive value and and at least one negative value somewhere in the interval $[0,b]$. (Notice that $g$ has a third option: $g$ can be zero throughout the interval. We won't have to think about that explicitly in what follows.)
• First we check that $g$ is continuous. We have: $f(x)$ is continuous (on its domain, $[0,b]$), the identity function, $x \mapsto x$, is continuous (on $[0,b]$) (It is, at least, a polynomial, which we should know is continuous.), and the difference of continuous functions is continuous. Therefore, $g$ is continuous (on $[0,b]$).
• Next we need to see that $g(x)$ takes positive and negative values. We don't have enough insight into what $f$ does to prove that directly, so we proceed by contradiction.
  • Suppose $g(x) \geq 0$ for all $x \in [0,b]$. That is, suppose that there is no point in $[0,b]$ where $g$ takes a negative value. Then, in particular, $g(b) \geq 0$, which forces $f(b) \geq b$. Since the image of $f$ is upper bounded by $b$, we must have $f(b) = b$. (The easiest way to see this is to draw the square $[0,b]\times[0,b]$, draw the diagonal from lower-left to upper-right (the $f(x) = x$ line). Then, try to draw the graph of a function starting anywhere on the left edge and ending anywhere on the right edge, without ever intersecting the diagonal. It's easy enough to stay above the diagonal right up until the right edge, at which you have no choice.) Therefore, either $c = b$ and the result holds or $g$ takes a negative value at least once on $[0,b]$.
  • Alternatively, suppose $g(x) \leq 0$ everywhere. Then, in particular, $g(0) \leq 0$. This forces $f(x) \leq 0$, which either forces $f(0) = 0$ and the result holds or $g$ takes a positive value at least once on $[0,b]$.

So that's how we arrive at how we work this problem. When we write it up, we should shuffle these around a little.

• Start with: $f(c) = c$ means we should study $g$.
• Show that $g$ is continuous.
• Show that $g$ takes at least one positive and at least one negative value.
• Apply the IVT.