# Math Help - If f is bounded on every neighborhood of x in [a,b], then f is bounded on [a,b]

1. ## If f is bounded on every neighborhood of x in [a,b], then f is bounded on [a,b]

$I = [a,b]$ and $f : I \to R$ such that
$\forall x \in I$, $f$ is bounded on a neighborhood $V_{\delta_{x}}(x)$ of x.

$J = (a,b)$ and $g : J \to R$ such that
$\forall x \in J$, $g$ is bounded on a neighborhood $V_{\delta_{x}}(x)$ of x.

Prove $f$ is bounded on $I$.
Give an example of a function $g$ that isn't bounded on $J$.

I have this question, and I don't know what $\delta_{x}$ means. I see that the endpoints satisfying the boundedness statement is the critical difference between f and g, making f bounded on $(a,b)$, but not requiring g to be. I have only worked with bounded continuous functions, and these functions aren't necessarily continuous.

2. Originally Posted by anon2194
$I = [a,b]$ and $f : I \to R$ such that
$\forall x \in I$, $f$ is bounded on a neighborhood $V_{\delta_{x}}(x)$ of x.

$J = (a,b)$ and $g : J \to R$ such that
$\forall x \in J$, $g$ is bounded on a neighborhood $V_{\delta_{x}}(x)$ of x.

Prove $f$ is bounded on $I$.
Give an example of a function $g$ that isn't bounded on $J$.

I have this question, and I don't know what $\delta_{x}$ means. I see that the endpoints satisfying the boundedness statement is the critical difference between f and g, making f bounded on $(a,b)$, but not requiring g to be. I have only worked with bounded continuous functions, and these functions aren't necessarily continuous.

Most probably they mean $V_{\delta_x}=(x-\delta_x,x+\delta_x)\,,\,\,s.t.\,\,V_{\delta_x}\su bset [a,b]\,\,or\,\,(a,b)$

== Since $[a,b]$ is compact there's a finite number of points $x_1,...,x_n\in [a,b]\,\,s.t.\,\,[a,b]=\bigcup\limits_{i=1}^nV_{\delta_{x_i}}$ , and since $\forall\,\, 1\le i\le\,,\, |f(x_i)|\le M_i$ , check now that $|f(x)|\le M:=max\{M_1,...,M_n\}\,,\,\forall x\in [a,b]$

== Counterexample: $g(x)=\frac{1}{x}\,\, in \,\,(0,1)$

Tonio

3. For 2 what about the function $g0,1)\to\mathbb{R}" alt="g0,1)\to\mathbb{R}" />, $g(x)=1/x$?

4. Originally Posted by tonio
== Since $[a,b]$ is compact there's a finite number of points $x_1,...,x_n\in [a,b]\,\,s.t.\,\,[a,b]=\bigcup\limits_{i=1}^nV_{\delta_{x_i}}$ , and since $\forall\,\, 1\le i\le\,,\, |f(x_i)|\le M_i$ , check now that $|f(x)|\le M:=max\{M_1,...,M_n\}\,,\,\forall x\in [a,b]$
Yes, I thought of dividing the closed interval and taking the supremum of the set of M's, but it seems like it's a naive approach to the problem. I know that it works, but it isn't using any theorems or techniques that I have learned. Just the definition of a neighborhood and a bounded function.

This approach also doesn't flow smoothly with the counter example $g(x)=\frac{1}{x}$, which I strangely thought of as well. I know it's intuitively obvious, but with the previous approach how do you go about showing that it is a counterexample.

5. Originally Posted by anon2194
Yes, I thought of dividing the closed interval and taking the supremum of the set of M's, but it seems like it's a naive approach to the problem. I know that it works, but it isn't using any theorems or techniques that I have learned. Just the definition of a neighborhood and a bounded function.

This approach also doesn't flow smoothly with the counter example $g(x)=\frac{1}{x}$, which I strangely thought of as well. I know it's intuitively obvious, but with the previous approach how do you go about showing that it is a counterexample.

Well, you have to use quite some important theorems to prove both things! For the first one you have to know that subset of the real lines is compact iff it is bounded and closed and thus [0,1] qualifies as such.
For the second one you have to use that a continuous function on a closed bounded interval is bounded there...

Both cases are pretty easy to understand and work very fine, but you have to use some rather important, non-trivial theorems.

Tonio

6. I haven't read about compact sets yet, which is why I felt like I was waving my hands. Thanks.

Yeah, the Heine-Borel Theorem.