Janet Forbes

Answered

2022-07-14

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

I am trying to use intermediate value theorem for continuous function. Here $|f(x)|$ is maximum with respect to $|f(y)|.$ Hence

$|f(y)|\le 1/2(|f(x)|+|f(y)|)\le |f(x)|.$

Now I can say $f(c)=1/2(|f(x)|+|f(y)|$ for some $c\in (a,b).$ But how to show $f(c)=0?$

I am trying to use intermediate value theorem for continuous function. Here $|f(x)|$ is maximum with respect to $|f(y)|.$ Hence

$|f(y)|\le 1/2(|f(x)|+|f(y)|)\le |f(x)|.$

Now I can say $f(c)=1/2(|f(x)|+|f(y)|$ for some $c\in (a,b).$ But how to show $f(c)=0?$

Answer & Explanation

Jayvion Mclaughlin

Expert

2022-07-15Added 14 answers

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

Now we get inductively a sequence $({x}_{n})$ in $[a,b]$ with

(*) $|f({x}_{n})|\le \frac{1}{{2}^{n}}|f({x}_{0})|.$

$({x}_{n})$ contains a convergent subsequence $({x}_{{n}_{k}})$ with limit $c\in [a,b]$.

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

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

Consequence: $f(c)=0$

Now we get inductively a sequence $({x}_{n})$ in $[a,b]$ with

(*) $|f({x}_{n})|\le \frac{1}{{2}^{n}}|f({x}_{0})|.$

$({x}_{n})$ contains a convergent subsequence $({x}_{{n}_{k}})$ with limit $c\in [a,b]$.

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

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

Consequence: $f(c)=0$

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