# Show that there exists $c\in[a,b]$ such that $f(c)=0$.

Question: Let $$f:[a,b]\to\mathbb{R}$$ be a continuous function with the property that for every $$x\in[a,b]$$, there exists $$y\in[a,b]$$ such that $$|f(y)|\le\frac{1}{2}|f(x)|$$. Show that there exists $$c\in[a,b]$$ such that $$f(c)=0$$.

Solution: Select any $$x\in [a,b].$$ Let $$x=x_1$$. Now by our hypothesis there exists $$x_2\in [a,b]$$ such that $$|f(x_2)|\le \frac{1}{2}|f(x_1)|.$$ Again by our hypothesis there exists $$x_3\in[a,b]$$ such that $$|f(x_3)|\le \frac{1}{2}|f(x_2)|\le \frac{1}{4}|f(x_1)|.$$ Continuing like this we will end up having a sequence $$(x_n)_{n\ge 1}$$ such that $$|f(x_n)|\le \frac{1}{2^{n-1}}|f(x_1)|, \forall n\in\mathbb{N}.$$ Notice that this implies that $$-\frac{1}{2^{n-1}}|f(x_1)|\le f(x_n)\le \frac{1}{2^{n-1}}|f(x_1)|, \forall n\in\mathbb{N}.$$ Thus by Sandwich theorem we can conclude that the sequence $$f(x_n)$$ is convergent and it converges to $$0$$.

Next notice that the sequence $$(x_n)_{n\ge 1}$$ is bounded. Thus, by Bolzano-Weierstrass theorem we can conclude that $$(x_n)_{n\ge 1}$$ has a convergent subsequence $$(x_{n_k})_{k\ge 1}$$. Let us assume that $$(x_{n_k})_{k\ge 1}$$ converges to $$c$$. Note that $$a\le c\le b$$. Now since $$f$$ is continuous on $$[a,b]$$, implies that $$f$$ is continuous at $$c$$. Thus by the sequential definition of limit we can conclude that $$f(x_{n_k})$$ converges to $$f(c)$$.

Now note that we have already shown that the sequence $$f(x_n)$$ converges to $$0$$, which implies that the subsequence $$f(x_{n_k})$$ also converges to $$0$$. This implies that $$f(c)=0.$$ This completes the proof.

It is also easy to see that if the inequality $$|f(y)|\le \frac{1}{2}|f(x)|$$ was replaced by the inequality $$|f(y)|\le \lambda |f(x)|$$ where $$|\lambda|<1$$ is arbitrary then also the statement in the question holds true.

Is this solution correct and rigorous enough and is there any other way to solve this problem?

• Yes it's correct. Aug 26 '20 at 18:08
• Looks good.$\mbox{}$ Aug 26 '20 at 18:11
• Great! Nice question as well. Aug 26 '20 at 18:32
• If you have some topology at your disposal, you can make it a bit shorter. Note that the infimum is a minimum for continuous functions on compact sets Aug 26 '20 at 18:34

## 1 Answer

Another way, maybe easier, to solve the problem :

Since $$|f|$$ is continuous on $$[a,b]$$, it attains its minimum, i.e. there exists $$x \in [a,b]$$ such that $$|f(x)| = \min_{t \in [a,b]} |f(t)|$$

Applying the property, there must exists $$y \in [a,b]$$ such that $$|f(y)| \leq \frac{1}{2}|f(x)|$$

If $$f(x) \neq 0$$, this contradicts the fact that $$|f(x)|$$ is the minimum of $$|f|$$. So $$|f(x)|=0$$.

• Wow. Such a clever way.+1. Aug 26 '20 at 18:41