# Thread: analysis - continuous functions

1. ## analysis - continuous functions

Just going through past papers as my anaysis exam is on Weds, but I'm quite stuck/unsure about 2 questions I've come across...

If g:[0,1] ---> [0,1] is a continuous function, show that there exists x in [0,1] such that g(x) = x.

Let F(x) = g(x) - x be defined on [0,1].
F is continuous on [0,1] since it is the difference of 2 continuous functions.
If F(0) = 0 or if F(1) =0 then we have our fixed point.
Otherwise, F(0) = g(0) - 0 > 0.
F(1) = g(1) - 1 < 0.

Since g(x) in [0,1] for all x in [0,1], by IVT there exists c in [0,1] such that F(c) = 0, hence g(c) = c.

Is this an OK answer, and would it be worthy of 8 marks?

Let f: R ---> R be a continuous function, such that f(x) in the rationals for all x. Show that f is a constant function.

This one I'm really stuck on and although it seems alot easier I have no idea what to say for it...can anyone please help?

2. Hi
Originally Posted by hunkydory19
Since g(x) in [0,1] for all x in [0,1], by IVT there exists c in [0,1] such that F(c) = 0, hence g(c) = c.
I don't understand why you write "Since g(x) in [0,1] for all x in [0,1]". Saying " $F(0)F(1)<0$ and $F$ is continuous on $[0,1]$ hence by IVT..." would be enough.

Let f: R ---> R be a continuous function, such that f(x) in the rationals for all x. Show that f is a constant function.

This one I'm really stuck on and although it seems alot easier I have no idea what to say for it...can anyone please help?
One possibility : assume that $f$ is not a constant function i.e. there exists $x_0 \in \mathbb{R}$ such that $f(x_0)\neq f(1)$ and then try to show that it is nonsense using the fact that $\mathbb{Q}$ is a dense subset of $\mathbb{R}$.

3. Originally Posted by hunkydory19
Just going through past papers as my anaysis exam is on Weds, but I'm quite stuck/unsure about 2 questions I've come across...

If g:[0,1] ---> [0.1] is a continuous function, show that there exists x in [0,1] such that g(x) = x.

Let F(x) = g(x) - x be defined on [0,1].
F is continuous on [0,1] since it is the difference of 2 continuous functions.
If F(0) = 0 or if F(1) =0 then we have our fixed point.
Otherwise, F(0) = g(0) - 0 > 0.
F(1) = g(1) - 1 < 0.

Since g(x) in [0,1] for all x in [0,1], by IVT there exists c in [0,1] such that F(c) = 0, hence g(c) = c.

Is this an OK answer, and would it be worthy of 8 marks?
[snip]
I'd be inclined to set it out in the following way:

Let f(x) = g(x) - x be defined on [0,1].

f is continuous on [0,1] since it is the difference of two continuous functions.

$0 \leq g(x) \leq 1 \Rightarrow -x \leq g(x) - x \leq 1 - x \Rightarrow -x \leq f(x) \leq 1 - x$.

Therefore $0 \leq f(0) \leq 1$ and $-1 \leq f(1) \leq 0$.

Therefore $f(0) \geq 0$ and $f(1) \leq 0$.

Therefore, by the IVT there exists c in [0, 1] such that $f(c) = 0 \Rightarrow g(c) - c = 0 \Rightarrow g(c) = c$.

4. Originally Posted by mr fantastic
I'd be inclined to set it out in the following way:

Let f(x) = g(x) - x be defined on [0,1].

f is continuous on [0,1] since it is the difference of two continuous functions.

$0 \leq g(x) \leq 1 \Rightarrow -x \leq g(x) - x \leq 1 - x \Rightarrow -x \leq f(x) \leq 1 - x$.

Therefore $0 \leq f(0) \leq 1$ and $-1 \leq f(1) \leq 0$.

Therefore $f(0) \geq 0$ and $f(1) \leq 0$.

Therefore, by the IVT there exists c in [0, 1] such that $f(c) = 0 \Rightarrow g(c) - c = 0 \Rightarrow g(c) = c$.
Now you should prove that any continuous bounded function over the real numbers f(x) must cross the function g(x) = x.

5. Cheers mr fantastic, that does indeed seem like a better answer.

By any chance do you have an hints/blatent help, for the second question...as i tried the approach flying squirrel suggested but couldnt get anywhere...

Thanks again!

6. Originally Posted by hunkydory19
as i tried the approach flying squirrel suggested but couldnt get anywhere...
Assume $f$ is not constant on $\mathbb{R}$ : there exists $x_0\in \mathbb{R}$ such that $f(x_0)\neq f(1)$.
The idea is to apply the definition of continuity at $x_0$ : $\forall \varepsilon >0, \, \exists\, \delta >0 \, | \, | x-x_0 | < \delta \Rightarrow | f(x)-f(x_0) | <\varepsilon$
As $\mathbb{Q}$ is a dense subset of $\mathbb{R}$, given $\delta>0$, there exists a rational number $q\,|\, |x_0-q|<\delta$ hence by choosing $\varepsilon=\frac{| f(1)-f(x_0)|}{2}$ we get that : $| f(q)-f(x_0) | = | f(1)-f(x_0) | < \frac{| f(1)-f(x_0)|}{2}$ which is a contradiction. Hence $f$ is constant on $\mathbb{R}$.

7. Originally Posted by hunkydory19
Let f: R ---> R be a continuous function, such that f(x) in the rationals for all x. Show that f is a constant function.
Using the IVT this proof is almost trivial.
If f is not constant then there two values $f\left( {x_1 } \right) \ne f\left( {x_2 } \right)$.
Between any two numbers there is an irrational number.
So some number between $x_1 \,\& \,x_2$ is mapped onto it by IVT.