Proving a function approaches 0 given certain conditions

**redsoxfan325** · April 14th 2009, 10:48 AM

Problem:

If $f(x)$ is a non-negative, Riemann-integrable, uniformly continuous function on $[0,\infty)$ , prove that $\lim_{x\to\infty}f(x) = 0$ .

Here's what I have so far.

I want to try to prove that $\sum_{n=1}^{\infty} f(n)$ converges. Since $f$ is uniformly continuous, $\forall~ \epsilon>0, \exists~ \delta>0$ such that $|f(x_i)-f(x_{i-1})|<\epsilon$ when $|x_i-x_{i-1}|<\delta$ .

Let $P$ be a partition of $[0,\infty)$ such that $\Delta x_i < \delta$ . Let $m_i = \inf f(x)$ on $[x_{i-1},x_i]$ and let $M_i = \sup f(x)$ on $[x_{i-1},x_i]$ . We know that:

$0 < \sum_{i=1}^{\infty} m_i\Delta x_i < \sum_{i=1}^{\infty} m_i\delta$ $\leq \int_0^{\infty}f(x)\,dx \leq \sum_{i=1}^{\infty} M_i\Delta x_i$ $< \sum_{i=1}^{\infty} M_i\delta$

Now I get stuck. I was thinking that since I know $|f(x_i)-f(x_{i-1})|<\epsilon$ , I could put some restrictions on $m_i$ and $M_i$ , but I'm not sure how to do this. Am I on the right track? Regardless, can someone give me a hint as to the best way to proceed?

**running-gag** · April 14th 2009, 11:50 AM

Originally Posted by redsoxfan325

Problem:
Here's what I have so far.

I want to try to prove that $\sum_{n=1}^{\infty} f(n)$ converges. Since $f$ is uniformly continuous, $\forall~ \epsilon>0, \exists~ \delta>0$ such that $|f(x_i)-f(x_{i-1})|<\epsilon$ when $|x_i-x_{i-1}|<\delta$ .

Let $P$ be a partition of $[0,\infty)$ such that $\Delta x_i < \delta$ . Let $m_i = \inf f(x)$ on $[x_{i-1},x_i]$ and let $M_i = \sup f(x)$ on $[x_{i-1},x_i]$ . We know that:

$0 < \sum_{i=1}^{\infty} m_i\Delta x_i < \sum_{i=1}^{\infty} m_i\delta$ $\leq \int_0^{\infty}f(x)\,dx \leq \sum_{i=1}^{\infty} M_i\Delta x_i$ $< \sum_{i=1}^{\infty} M_i\delta$

Now I get stuck. I was thinking that since I know $|f(x_i)-f(x_{i-1})|<\epsilon$ , I could put some restrictions on $m_i$ and $M_i$ , but I'm not sure how to do this. Am I on the right track? Regardless, can someone give me a hint as to the best way to proceed?

Hi

I don't understand why you "want to try to prove that $\sum_{n=1}^{\infty} f(n)$ converges". This is an hypothesis since f is Riemann-integrable.

Suppose that $\lim_{x \rightarrow \infty} \neq 0$

Then f being positive $\lim_{x\rightarrow\infty} = L > 0$ , L being eventually infinite

I suppose that L is finite

$\forall~ \epsilon>0, \exists~ A>0$ such that $|f(x) -L|<\epsilon$ when $x > A$

Let $\epsilon = \frac{L}{2}$
Let A > 0 such that $\forall~ x, x > A \implies |f(x) - L|<\frac{L}{2}$
Then $\forall~ x, x > A \implies |f(x)|>\frac{L}{2}$

Let 's take the subdivision of $[0,\infty)$ $x_i = iA$ starting from i = 0
$\Delta x_i = A$

Let $m_i = \inf f(x)$ on $[x_i,x_{i+1}]$
Since $\forall i \geq 1 \:x_i \geq A$ then $m_i \geq \frac{L}{2} \:\forall i \geq 1$

$\int_0^{\infty}f(x)\,dx > \sum_{i=0}^{\infty} m_i\Delta x_i$

$\sum_{i=0}^{n} m_i\Delta x_i = A\: \sum_{i=0}^{n} m_i = A\: \left(m_0 + \sum_{i=1}^{n} m_i\right)$

Spoiler:

**redsoxfan325** · April 14th 2009, 03:06 PM

Where did you use the fact that f is uniformly continuous?

And I wanted to prove that $\sum_{n=1}^{\infty} f(n)$ converged because I thought that only followed if f is monotonically decreasing. There are unbounded functions that are integrable.

**manjohn12** · April 15th 2009, 07:31 AM

Originally Posted by redsoxfan325

Where did you use the fact that f is uniformly continuous?

And I wanted to prove that $\sum_{n=1}^{\infty} f(n)$ converged because I thought that only followed if f is monotonically decreasing. There are unbounded functions that are integrable.

Unbounded functions are not Riemann integrable.

**redsoxfan325** · April 15th 2009, 07:42 AM

Consider this function:

$f(x) = \left\{ \begin{array}{lr} 0 &: 0\leq x<1\\ -4n^7(x-n)(x-(n+n^{-3})) &: x\in[n,n+n^{-3}]\\ 0 &: otherwise \end{array} \right\}$

It's continuous (though not uniformly), positive, unbounded, and Riemann integrable.

I need to prove that if I add the condition that it's uniformly continuous, $\lim_{x\to\infty} f(x) = 0$ .

**redsoxfan325** · April 15th 2009, 08:44 AM

Oh, and runninggag, how do you know the limit L exists? What if you're looking at a function like $x\sin x$ , which I don't think has a limit as $x\to\infty$ .

**Opalg** · April 15th 2009, 12:27 PM

If $f(x)$ is a non-negative, Riemann-integrable, uniformly continuous function on $[0,\infty)$ , prove that $\lim_{x\to\infty}f(x)=0$ .

Careful! This is an improper Riemann integral. You cannot analyse a Riemann integral over an infinite interval by using partitions. You need to define it as the limit of integrals over finite intervals.

What you are told here is that $\int_0^\infty\!\!\!f(x)\,dx \mathrel{\mathop=^\text{d{}ef}} \lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx$ exists. This implies that $\int_X^Y\!\!\!f(x)\,dx$ can be made arbitrarily small by taking X and Y sufficiently large (I'll make that more precise later).

The uniform continuity of f means that given $\varepsilon>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$ .

Now suppose that $f(x)\geqslant2\varepsilon$ for some value of x. Then (by the triangle inequality) it will follow that $f(y)>\varepsilon$ whenever whenever $|x-y|<\delta$ . Therefore $\int_x^{x+\delta}\!\!\!f(y)\,dy>\varepsilon\delta$ . But the fact that $\lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx$ exists tells you that there exists $X_0$ such that $\int_X^Y\!\!\!f(x)\,dx<\varepsilon\delta$ whenever $Y>X>X_0$ (as in the previous paragraph). Therefore $f(x)\geqslant2\varepsilon\:\Rightarrow\:x<X_0$ . Equivalently, $f(x)<2\varepsilon$ whenever $x>X_0$ . This says that $f(x)\to0$ as $x\to\infty$ .

**redsoxfan325** · April 15th 2009, 07:17 PM

Originally Posted by Opalg

Now suppose that $f(x)\geqslant2\varepsilon$ for some value of x. Then (by the triangle inequality) it will follow that $f(y)>\varepsilon$ whenever whenever $|x-y|<\delta$ . Therefore $\int_x^{x+\delta}\!\!\!f(y)\,dy>\varepsilon\delta$ . But the fact that $\lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx$ exists tells you that there exists $X_0$ such that $\int_X^Y\!\!\!f(x)\,dx<\varepsilon\delta$ whenever $Y>X>X_0$ (as in the previous paragraph). Therefore $f(x)\geqslant2\varepsilon\:\Rightarrow\:x<X_0$ . Equivalently, $f(x)<2\varepsilon$ whenever $x>X_0$ . This says that $f(x)\to0$ as $x\to\infty$ .

Why should we prove that $f(x)\geqslant2\varepsilon \implies x<X_0$ ? Can't we just go straight to proving $f(x)<2\varepsilon \implies x>X_0$ or do we need to prove the former to prove the latter?

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