 Janet Forbes

2022-07-14

Let $f:\left[a,b\right]\to \text{R}$ be a continuous function such that for each $x\in \left[a,b\right]$ there exists $y\in \left[a,b\right]$ such that $|f\left(y\right)|\le \frac{1}{2}|f\left(x\right)|.$ Prove that there exists $c\in \left[a,b\right]$ such that $f\left(c\right)=0.$

I am trying to use intermediate value theorem for continuous function. Here $|f\left(x\right)|$ is maximum with respect to $|f\left(y\right)|.$ Hence
$|f\left(y\right)|\le 1/2\left(|f\left(x\right)|+|f\left(y\right)|\right)\le |f\left(x\right)|.$
Now I can say $f\left(c\right)=1/2\left(|f\left(x\right)|+|f\left(y\right)|$ for some $c\in \left(a,b\right).$ But how to show $f\left(c\right)=0?$ Jayvion Mclaughlin

Expert

Let ${x}_{0}\in \left[a,b\right]$. Then there is ${x}_{1}\in \left[a,b\right]$ such that $|f\left({x}_{1}\right)|\le \frac{1}{2}|f\left({x}_{0}\right)|.$

Now we get inductively a sequence $\left({x}_{n}\right)$ in $\left[a,b\right]$ with
(*) $|f\left({x}_{n}\right)|\le \frac{1}{{2}^{n}}|f\left({x}_{0}\right)|.$
$\left({x}_{n}\right)$ contains a convergent subsequence $\left({x}_{{n}_{k}}\right)$ with limit $c\in \left[a,b\right]$.

$f$ is continuous, thus $f\left({x}_{{n}_{k}}\right)\to f\left(c\right)$ .

From (*) we get: $f\left({x}_{{n}_{k}}\right)\to 0$.

Consequence: $f\left(c\right)=0$

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