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

<br /> \int_1^\infty |f(x)|dx &lt; \infty,<br />

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

<br /> f(x) = \begin{cases}<br /> 1, &amp; x \in [0,a],\\<br /> 0, &amp; \text{otherwise}.<br /> \end{cases}<br />

 

[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. ;))