Can anyone come up with a Lebesgue-integrable function that

[AxiomOfChoice]

...satisfies the following conditions:

(1) Is continuous on $[1,\infty)$, and

(2) Does not have a limit as $x\to \infty$.

Apparently, such a function $f(x)$ exists, but I cannot think of an example for the life of me. Remember: The function must also satisfy

$$\int_1^\infty |f(x)|dx < \infty,$$

where "$\int$" is the Lebesgue integral.

 

[Hurkyl]

Can you think of a Riemann-integrable function f such that:

• Max(f) = 1
• Min(f) = 0
• The integral over all of R is a (a is a previously chosen positive real number

 

[AxiomOfChoice]

Can you think of a Riemann-integrable function f such that:

• Max(f) = 1
• Min(f) = 0
• The integral over all of R is a (a is a previously chosen positive real number

Of course. How about

$$f(x) = \begin{cases} 1, & x \in [0,a],\\ 0, & \text{otherwise}. \end{cases}$$

 

[disregardthat]

On the line $$[1,\infty)$$, at each integer point n, draw an isoceles of height $$2^n$$ and with base width $$4^{-n}$$. Let the function be the curve of these isoceles when they occur, and 0 when they don't. This function is obviously continuous, and

$$\int^{\infty}_1 |f(x)| dx = \sum_{n=1}^{\infty} \frac{2^{n}4^{-n}}{2}= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2}$$

However, the function has no limit as $$x \to \infty$$. As an extra bonus it is not even bounded.

 

[Ben Niehoff]

Jarle, some of your triangles overlap, but the basic principle still works.

Is it possible to find a $C^\infty$ function that satisfies the OP's criteria? Yes, I see it is possible after just writing that...

How about an analytic function that satisfies the criteria? For that, I'm not sure...

 

[AxiomOfChoice]

Jarle, some of your triangles overlap...

I don't see why this is so. Can you give an example of two overlapping triangles?

Also, what do you mean by "OP's criteria?"

 

[AxiomOfChoice]

On the line $$[1,\infty)$$, at each integer point n, draw an isoceles of height $$2^n$$ and with base width $$4^{-n}$$. Let the function be the curve of these isoceles when they occur, and 0 when they don't. This function is obviously continuous, and

$$\int^{\infty}_1 |f(x)| dx = \sum_{n=1}^{\infty} \frac{2^{n}4^{-n}}{2}= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2}$$

However, the function has no limit as $$x \to \infty$$. As an extra bonus it is not even bounded.

Perfect. I feel like a fool for not having thought of this myself. Thanks!

 

[Hurkyl]

How about an analytic function that satisfies the criteria? For that, I'm not sure...

I feel like this is one of those questions that is either "obviously yes" or "obviously no", but I don't know which.

My first inclination is a function like

$$f(x) = \exp\left(\frac{x}{2} \log( \sin(x)^2 ) \right)$$​

which, on R, simplifies to

$$|\sin x|^{x}$$​

or maybe replace x/2 with something even faster growing.


I haven't ground through the analysis to see if this actually has a finite integral.


However, http://www.wolframalpha.com/input/?i=Integrate[Exp[Exp[x]/2+Log[Sin[x]^2]],+x]


Is that analytic? Well, this one is more obviously so: http://www.wolframalpha.com/input/?i=Integrate[Exp[Exp[x]/2+Log[(1/2)+++(1/2)Sin[x]^2]],+x]

 

[Mute]

I feel like this is one of those questions that is either "obviously yes" or "obviously no", but I don't know which.

My first inclination is a function like

$$f(x) = \exp\left(\frac{x}{2} \log( \sin(x)^2 ) \right)$$​

which, on R, simplifies to

$$|\sin x|^{x}$$​

or maybe replace x/2 with something even faster growing.I haven't ground through the analysis to see if this actually has a finite integral.However, http://www.wolframalpha.com/input/?i=Integrate[Exp[Exp[x]/2+Log[Sin[x]^2]],+x]Is that analytic? Well, this one is more obviously so: http://www.wolframalpha.com/input/?i=Integrate[Exp[Exp[x]/2+Log[(1/2)+++(1/2)Sin[x]^2]],+x]

How about replacing the triangles in the given solution to the original problem with Gaussians?

$$f(x) = \sum_{n=1}^\infty \frac{2^n}{\sqrt{2\pi}} \exp\left[-\frac{(x-n)^2}{2\sigma_n^2}\right]$$
where $\sigma_n = 4^{-n}$?

It's possible this isn't analytic since we're summing an infinite number of terms, but it'd be the first thing I'd try (if I felt like trying to prove/disprove things like analyticity). (Of course, even if analyticity is proved doing the resulting integral wouldn't be too fun - at least on the [1,infinity) interval. (-infinity,infinity) wouldn't be so bad. ;))