# Prove that there is a number $x_0 \in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\le x < x_0$.

Suppose that the function $f:[0,1]\rightarrow \mathbb{R}$ is continuous, $f(0)>0$, and $f(1)=0$. Prove that there is a number $x_0 \in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\le x < x_0$; that is, there is a smallest point in the interval $[0,1]$ at which the function $f$ attains the value $0$.

Since $f$ is continuous on a closed an bounded interval, $[0,1]$, then by the extreme value theorem, $f$ attains both a minimum and a maximum value. The minimum value will obviously be $0$, but I'm not really sure where to go after stating this. Any suggestions?

• I would do a proof by contradiction. – shade4159 Nov 6 '13 at 4:43
• @shade4159 I'm not really sure how I would show this by contradiction. If I suppose there is not a number $x_0\in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0\le x<x_0$, then can't there still be a minimum and maximum on the interval? – TheMobiusLoops Nov 6 '13 at 4:51
• Assume there is no smallest point in the interval $[0,1]$ at which $f$ attains the value $0$. So $\forall x_0 \in [0,1]| f(x_0) = 0, \exists x_1 \in [0,1]| f(x_1) = 0, x_1 < x_0$ – shade4159 Nov 6 '13 at 4:56

If there is no smallest root, then, for any $c > 0$, there is a root $r$ such that $0 < r < c$.

Use this to prove that $f$ is not continuous at zero.

The existence of a root is apparent in the region $(0,1]$ as 1 is a root. So, now that we know the roots exist, we can take the least root in the set $S:=\{x_i\mid f(x_i)=0\}$ and call that element $x_0$, since the set of roots of a continuous function is closed. Since, for all $x\in(0,x_0)$, $f(x)\not=0$, $f(0)=1$, $f(x_0)=0$, and $f$ is continuous, all values between $x_0$ and $0$ must be greater than 0.

For a proof by contradiction, assume: $$\forall x_0 \in [0,1] \,| \, f(x_0) = 0, \exists x_1 \in [0,1]\,|\, f(x_1) = 0, x_1 < x_0$$

Then, since we are given $f$ is continuous on $[0,1]$ and $f(0) > 0$, it should be relatively simple to show a contradiction exists.

Starting from the hypothesis that $f:[0,1]\rightarrow \mathbb{R}$ is a continuous function (on a compact set $[0,1]$) satisfying $f(0)>0$ and $f(1)=0$ one has in particular that there exist $c,d \in [0,1]$ such that $\displaystyle f(c)=\inf_{x\in [0,1]} f(x)$ and $\displaystyle f(d)=\sup_{x \in [0,1]}f(x)$.

In particular, one has the flag of inequalities $$f(c)\leq f(1)=0<f(0)\leq f(d).$$

The proof that $f(x_0)=0$ for some $0\leq x_0<1$ is a direct consequence for the intermediate value theorem for continuous functions. Indeed, if $f(c)=0$ then $x_0$ is a root of the equation $f(x)$ on the interval $[0,x_0]$. Otherwise, one has $f(c)<0<f(0)$ together with the continuity condition assures the existence of $x_0$ such that $f(x_0)=0$ fulfils on the open interval $(0,c)$.

The proof that the inequality $f(x)>0$ holds on the interval $0\leq x<x_0$ is a natural consequence of the continuity of the function on the point $0$.

In concrete, for every $\varepsilon>0$ it is possible to find a $\delta>0$ such that for every $x \in (-\delta,\delta)\cap [0,x_0)$ there holds

$$-\varepsilon<f(x)-f(0)<\varepsilon.$$

In particular, for the choice $\varepsilon=f(0)$ we then have that $f(x)>0$ in a neighborhood $(0,\delta)$ of the interval $[0,x_0)$ ($\delta\leq x_0$). This proves that $f(x)>0$ on $(0,x_0)$. Thereby $f(x)>0$ on $[0,x_0)$, as desired.

• The existence of $x_0$ with $f(x_0)=0$ is obvious as we may take $x_0=1$. The real challenge is to find $x_0$ such that $f(x_0)=0$ and $f(x)>0$ for $x\in[0,x_0)$. And your proof at the end only shows that $f(x) >0$ in $[0,\delta)$ and hence wrong. The right answer is given by considering set $A=\{x\mid f(x) =0\}$ and then using $x_0=\inf\,A$. – Paramanand Singh May 25 '18 at 8:11
• Yes! You are indeed right. The set $A=f^{-1}(\{0\})$ is closed. And due to continuity arguments on a compact set so that the existence of $x_0$ is guaranteed. – Nelson Faustino May 25 '18 at 15:01