Prove that there is a first time the function is zero

Question

Let $f : [0, 1] \to \mathbb R$ be continuous and satisfy $f(0) = −1$, $f(1) = 1$. Prove that there is a first time at which $ f $ is zero: that is, a number $s \in (0, 1)$ with $f(s) = 0$ but $f(t) \not= 0$ if $t < s$.

I'm thinking of using the least upper bound property on the set and somehow proving it is $ 0 $ at the value?

Any hints?

Answer 1

Suppose there is no first time. Then, for any $s\in (0,1)$ such that $f(s)=0$ (why do you know that even one such $s$ exists?) there exists $s'<s$ such that $f(s')=0$ as well. We can keep doing this and get a decreasing sequence of points $\{s_i\}_{i\in\mathbb{N}}$. Does this sequence converge? How do convergent sequences behave with continuous functions?

Answer 2

We need the following fact: Since $f$ is continuous, the set $Z(f)=\{x\in[0,1]:f(x)=0\}$ is a closed set.

It is non-empty by intermediate value theorem. Then $s=\inf Z(f)=\min Z(f)$ is the first time at which $f$ is zero. It is clear $s\in (0,1)$.

Answer 3

We know $f(c) = 0$ for some $c \in [0,1]$ by the Intermediate Value Theorem.

So $S = \{x|x \in [0,1]; f(x) = 0\}$ is not an empty set and it is bounded by $0, 1$. So $a=\inf S$ exists.

What is $f(a)$.

Well $f(a) > 0$ is impossible because $f(0) < 0$ so by IVT there would be an $d; 0 < d < a$ so that $f(0)=0$. So $d \in S$ but $d < \inf S$ which is impossible.

And $f(a)< 0$ is impossible because $f$ is continuous, and if we let $\epsilon = |f(a)| > 0$ we can find a $\delta > 0$ so the for all $c; a-\delta < c < a+\delta$ is well be true that $|f(c)- f(a)| < \epsilon$. But that means $f(a) < f(c) < 0$ so $f(c)< 0$ for all such $c$. So if $d \in S$ then $d \ge a+\delta$ and $a + \delta$ is a lower bound of $S$, contradicting that $a = \inf S$.

So $f(a) = 0$ and no other value below is such.

Answer 4

For an argument from first principles, try defining $$ S = \{t \in (0, 1) : f(x) < 0 \text{ for all } x \in [0, t) \}. $$ By the continuity of $f$ at $0$, $S$ is non-empty. By the continuity of $f$ at $1$, $S$ has an upper bound in $(0, 1)$. So $S$ has a least upper bound, $s \in (0, 1)$. Now prove (i) $s \in S$, (ii) $f(s) \leqslant 0$, (iii) $f(s) \geqslant 0$.

(I hope this is OK. You asked for a "hint", so I've left plenty of details to fill in. It's possible I've fooled myself!)