existence of fixed points under a continuous map.

Question

Let $f: [0, \infty) \to [0, \infty)$ be a continuous function. Which of the following is correct?

1. $\exists a \in [0, \infty)$ such that $f(a) = a$.

2. If $f(x) \leq M \forall x \in [0, \infty)$ for some $M > 0$, then $\exists a \in [0, \infty)$ such that $f(a) = a$

3. If $f$ has a fixed point , then it must be unique.

4. $f$ does not have a fixed point unless it is Differentiable on $(0, \infty)$

My Attempt : For 1, If I take $f(x) = e^x$ which is continuous but it does not have any fixed point . For 3, if I take $f(x) = x$ which is continuous but it has uncountable fixed points. I have no idea about other options. Please help me.

Answer 1

1. $f(x)=e^x$ is a great counterexample.

2. Is true, see below.

3. $f(x)=x$ is the perfect counterexample.

4. We can construct a counterexample consisting of multiple straight lines (of different slopes, but joined together for continuity) that crosses the quadrant bisector many times, thus having many fixed points dispite not being differentiable. in the picture below, the green dotted line is $y=x$. Each point on this line is a fixed point.

For sratement 2. Let $f(x)$ be a continuous, bounded function, and $M$ the upper bound. Then $f(x)\leq M$ for all $x\in[0, +\infty)$.

If $f(0)=0$, we found a fixed point and we are done. If not, $f(0)>0$, then let consider the function $g(x)=f(x)-x$. Now $$g(0)=f(0)-0>0$$ and for $a>M$, $$g(a)=f(a)-a\leq M-a<0$$ By the intermediate value theorem, there exist a $c \in (0, a)$ such that $g(c)=0$. $$0=g(c)=f(c)-c \implies f(c)=c$$ We found a fixed point.

Answer 2

For 4) consider $f(x)=1+|x-1|$. Note that $x=1$ is a fixed point but $f$ is not differentiable at $1$.

Answer for 2): Let $g(x)=x-f(x)$. Then $g(0) \leq 0$ and $g(x) >0$ for $x>M$. By continuity there exists a point $a$ such that $g(a)=0$. This gives $f(a)=a$.