Suppose $\displaystyle f:\mathbb{R} \longrightarrow [0,\infty) $ is continuous and that $\displaystyle \int f < \infty$. Is it always true that $\displaystyle \lim_{x\to\infty}f(x)=0?$

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- Apr 16th 2010, 12:12 PMdavidmccormickFlat tail?
Suppose $\displaystyle f:\mathbb{R} \longrightarrow [0,\infty) $ is continuous and that $\displaystyle \int f < \infty$. Is it always true that $\displaystyle \lim_{x\to\infty}f(x)=0?$

- Apr 16th 2010, 12:17 PMDrexel28
Clearly we must have that eventually $\displaystyle f(x)$ is non-increasing. So, let $\displaystyle f(x)$ be non-increasing for $\displaystyle x\geqslant N\in\mathbb{N}$. Then, $\displaystyle f$ is continuous non-negative non-increasing function. Thus, $\displaystyle \int_{N}^{\infty}f<\infty\implies \sum_{n=N}^{\infty}f(n)<\infty$ and so $\displaystyle \lim_{n\to\infty}f(n)=\lim_{x\to\infty}f(x)=0$.

- Apr 16th 2010, 01:56 PMLaurent
What do you mean?

In general, functions $\displaystyle f\geq 0$ such that $\displaystyle \int f<\infty$ are not non-increasing. Some of them converge to 0, others don't.

Here's an example. Let me describe the graph of $\displaystyle f$. It consists of triangular peaks that get thinner and higher : around $\displaystyle n$, we put a triangular peak of width $\displaystyle \frac{1}{n^3}$ and height $\displaystyle n$. (Is it clear?)

The area of the triangle at $\displaystyle n$ is $\displaystyle \frac{1}{n^2}$. Since $\displaystyle \sum_n \frac{1}{n^2}<\infty$, the total area under the curve is finite, i.e. $\displaystyle \int f(x)dx<\infty$. However, $\displaystyle f$ clearly does not converge to 0. We even have $\displaystyle f(n)=n\to +\infty$ for $\displaystyle n\in\mathbb{N},n\to\infty$. - Apr 16th 2010, 02:00 PMDrexel28