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Use the Boundedness Theorem to show that if the function $f: [0,1] \rightarrow \mathbb{R}$ is continuous and $f(x) \ne 0$ for all $x \in [0,1]$, then there exists $\delta > 0$ such that $|f(x)| > \delta$ for all $x \in [0,1]$.

The Boundedness Theorem is stated as such in our notes: A continuous function on a closed bounded interval is bounded and attains its bounds.

I understand what the question is saying, I just have no idea how to answer it.

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2 Answers 2

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"It attains its bounds" means that there is a minimum value that $|f(x)|$ attains. Because $f$ is continuous, we know that $|f|$ is continuous. Hence $|f|$ attains minimum and maximum values.

As $f(x)\neq 0$ for any $x$, this means that $|f(x)|\neq 0$ for any $x$. Thus $\inf|f(x)| = \min|f(x)| > 0$. Now you can just set $\delta = \min|f(x)|/2$.

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  • $\begingroup$ Would I have to prove anything further or would this suffice as my solution. i.e. Setting, δ=min|f(x)|/2, we have shown there exists δ>0 s.t. |f(x)|>δ for all x∈[0,1]. This seems okay in my head but surely it can't be that easy! lol $\endgroup$
    – user499701
    Nov 5, 2017 at 23:33
  • $\begingroup$ If you want to be more explicit, you can say consider $s = \max\{f(x) : f(x) < 0\} = \sup\{f(x) : f(x) < 0\}$ and $t = \min\{f(x): f(x) > 0\} = \inf\{f(x): f(x) > 0\}$. Then $\min|f(x)| = \min\{|s|, |t|\}$. The "boundedness theorem" tells you that the infimum is in fact a minimum and the supremum is in fact a maximum, so they must be non-zero. I don't think you need to write anything more than I did in the answer though. $\endgroup$
    – Harambe
    Nov 5, 2017 at 23:40
  • $\begingroup$ I think the answer is better without the previous comment, as it is not clear why the infimum of the restricted function would be attained. It’s simpler to say that “absolute value” is continuous and composition of continuous functions is continuous. $\endgroup$
    – user334639
    Nov 5, 2017 at 23:46
  • $\begingroup$ Yeah I was just editing my answer as you made that comment :) $\endgroup$
    – Harambe
    Nov 5, 2017 at 23:48
  • $\begingroup$ You need intermediate value theorem to prove that $f(x)\neq 0$, right? $\endgroup$
    – Ennar
    Nov 5, 2017 at 23:54
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If you are interested it can be proved by an argument of absurdity too:

Let assume the negated statement: " $\forall \delta>0, \exists x\in[0,1]\text{ s.t. } |f(x)|\le \delta$ "

In particular for any $n\in\mathbb N$ there exists $x_n\in[0,1]$ such that $|f(x_n)|\le \frac 1n$.

$(x_n)_n$ is a sequence in $[0,1]$ compact, so we can extract a convergent subsequence $x_{\phi(n)}\to a$ with $a\in[0,1]\quad$ ($\phi\nearrow$ strict.).

Since $f$ is continuous then $f(a)=0$

which contradicts with the requirement that $f(x)\neq 0,\forall x\in[0,1]$.

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