# Can I get some examples of a function with following properies

1. Apr 21, 2014

a function, say f, can be integrated m times (say m less than infinity) AND its first derivative is bounded and continuous. could someone provide some examples of this?

i'm trying to show f(x) will go to 0 as x-> infinity but i think if i could think of some examples, it might be a good place to start and/or get a feel of this problem.

we're dealing in Reals btw, R+

2. Apr 21, 2014

### disregardthat

First, if a function is (Riemann) integrable, then any antiderivative is differentiable hence continuous, and therefore integrable. So you don't need the condition to be m times integrable, only once.

Second, you require it to be differentiable. Then by the same token it is integrable. Your conditions may be shortened to:

f is differentiable and its derivative is continuous and bounded. Any linear function is an example of this.

Thus it's not the case that f(x) approaches 0 as x approaches infinity.

3. Apr 21, 2014

### mathwonk

when you say "integrated" do you mean "anti differentiated"? and when you say m times, do you mean no more than m times? I am a little puzzled by your question. If you start from any bounded continuous function and integrate it (not anti differentiate) once you get a function with bounded continuous derivative that can be integrated (and anti differentiated) a many times as you like. but it does not necessarily go to 0 at infinity it seems to me. On the other hand any smooth "bump function" with bounded support also has all your properties and does go to zero at infinity.

4. Apr 21, 2014

i may have oversimplified the problem then because the statement i made in OP is the one that needs to be proved.

the actual question is

f(x) belongs to C'_b and L_p space. Need to show that f(x) -> 0 as x-> 0

this is L_p space http://en.wikipedia.org/wiki/Lp_space
basically anything in L_p can be integrated p times

f(x) in C'_b means it's first deriv is continuous and is bounded

5. Apr 21, 2014

### pwsnafu

No, that's not what it means. "Integrate p times" means that you have p number of integrals. Lp means f raised to the pth power is integrable.

6. Apr 21, 2014

### micromass

Staff Emeritus
That's not really what it means. It means that $\int |f|^p$ is finite.

Assume $f\geq 0$. Suppose $\lim_{x\rightarrow +\infty} f(x) \neq 0$. Show that there then exists an $\varepsilon>0$ and a sequence $x_n\rightarrow +\infty$ such that $f(x_n)>\varepsilon$.

By mean-value theorem, we have $|f(x) - f(y)|\leq M|x-y|$ for some bound $M$. Thus for all $y\in [x-\delta,x+\delta]$, we have $|f(x) - f(y)|\leq 2\delta M$.

Let $I_n = [x_n -\delta,x_n+\delta]$, can you find a lower bound for

$$\int_{I_n} f(x)^p$$

Does that imply your result?

7. Apr 22, 2014

ah that makes sense. i've couple of questions on what you did above and i hope you don't mind. it has been many years since i took analysis course so it seems i've forgotten what might be obvious.

how is it that $y\in [x-\delta,x+\delta]$? i mean how is it that a variable in y belongs to interval in x? shouldn't it be like $y\in [y_1,y_2]$? i'm sure you are correct with what you said but i'm trying to figure out how this works.

I vaguely recall there being some kind of method for finding upper/lower bound of integrals. basically that yeah?

once i find the lower bound i assume there's a theorem that leads me the result i want or somewhere close to it.

eh i need to buy a analysis book.

Last edited: Apr 22, 2014
8. Apr 22, 2014

okay so upper bound of that integral is $M(x_n+\delta-x_n+\delta)=2M\delta$ while lower bound is $2m\delta$ for

$m\leq f(x) \leq M$

Last edited: Apr 22, 2014
9. Apr 22, 2014

### mathwonk

if a graph has finite area over an infinite base it seems plausible that the height goes to zero, no?

10. Apr 22, 2014

### gopher_p

It seems like it should be true, but it's not.

11. Apr 23, 2014

### disregardthat

If the first derivative is bounded and continuous, I think it's the case.

12. Apr 23, 2014

could you shed some light on why you think this is the chase? how does an additional requirement of "first deriv bounded and cont" make height f(x) approach 0.

I know that just because $f(x) \in L_p$ space it does not mean $f(x) -> 0$ as $x -> \inf$. I know for sure that the additional requirement of $f(x) \in (C_b)^'$ is needed.

Last edited: Apr 23, 2014
13. Apr 23, 2014

why is this not true? i know adding "f(x) is bounded and cont makes it true" (this is what i'm trying to prove) but i'm curious why not having that requirement makes it false. any counter example?

14. Apr 23, 2014

### disregardthat

Consider a step function on the set $A = \bigcup_{n = 1}^{\infty} [n-\frac{1}{2^n},n+\frac{1}{2^n}]$. The integral of this function is the sum $\sum_{n=1} 2^{-n+1} = \frac{1}{2}$. The maximal height remains 1 on any interval $[x,\infty)$. You can easily find a smooth function behaving in a similar way, with maximal height 1 on each such interval and converging integral. But any smooth function constructed in this manner will have unbounded derivatives.

If the derivative is bounded, try to find a function composed of congruent isoceles triangles (with top points $\epsilon$ in the x-values of the sequence $x_n$ micromass defined) strictly smaller than |f(x)|, with diverging integral. This may be found since you have a strict limit of the derivative of f(x) and thus how fast f(x) increases and decreases near each point $x_n$. You probably need to consider a subsequence of $x_n$ such that the triangles are disjoint.

Last edited: Apr 23, 2014
15. Oct 15, 2014

### platetheduke

Let $f\in L^{p}$ have a continuous, bounded first derivative. Note that since $f\in L^p$, $\epsilon m(\{x:\vert f(x)\vert\geq\epsilon^{1/p}\})\leq\int_{\{x:\vert f(x)\vert\geq\epsilon^{1/p}\}}\vert f\vert^pdm<\infty$ for every $\epsilon>0$. You know that $f^p$ stays below $\epsilon$ on all but a set of finite measure. If this set is unbounded, use the triangle idea of disregardthat to show that the integral must be infinite, which contradicts the fact that $f$ is in $L^p$.