# Thread: Proving a function approaches 0 given certain conditions

1. ## Proving a function approaches 0 given certain conditions

Problem:
If $\displaystyle f(x)$ is a non-negative, Riemann-integrable, uniformly continuous function on $\displaystyle [0,\infty)$, prove that $\displaystyle \lim_{x\to\infty}f(x) = 0$.
Here's what I have so far.

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

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

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

Now I get stuck. I was thinking that since I know $\displaystyle |f(x_i)-f(x_{i-1})|<\epsilon$, I could put some restrictions on $\displaystyle m_i$ and $\displaystyle M_i$, 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?

2. Originally Posted by redsoxfan325
Problem:
Here's what I have so far.

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

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

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

Now I get stuck. I was thinking that since I know $\displaystyle |f(x_i)-f(x_{i-1})|<\epsilon$, I could put some restrictions on $\displaystyle m_i$ and $\displaystyle M_i$, 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 $\displaystyle \sum_{n=1}^{\infty} f(n)$ converges". This is an hypothesis since f is Riemann-integrable.

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

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

I suppose that L is finite

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

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

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

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

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

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

Spoiler:

But
$\displaystyle \sum_{i=1}^{n} m_i \geq n \:\frac{L}{2}$

Therefore
$\displaystyle \sum_{i=0}^{n} m_i\Delta x_i \geq A\: \left(m_0 + n\:\frac{L}{2}\right)$

And
$\displaystyle \sum_{i=0}^{\infty} m_i\Delta x_i = +\infty$

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

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

4. Originally Posted by redsoxfan325
Where did you use the fact that f is uniformly continuous?

And I wanted to prove that $\displaystyle \sum_{n=1}^{\infty} f(n)$ 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.

5. Consider this function:

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

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, $\displaystyle \lim_{x\to\infty} f(x) = 0$.

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

7. If $\displaystyle f(x)$ is a non-negative, Riemann-integrable, uniformly continuous function on $\displaystyle [0,\infty)$, prove that $\displaystyle \lim_{x\to\infty}f(x)=0$.
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 $\displaystyle \int_0^\infty\!\!\!f(x)\,dx \mathrel{\mathop=^\text{d{}ef}} \lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx$ exists. This implies that $\displaystyle \int_X^Y\!\!\!f(x)\,dx$ 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 $\displaystyle \varepsilon>0$ there exists $\displaystyle \delta>0$ such that $\displaystyle |f(x)-f(y)|<\varepsilon$ whenever $\displaystyle |x-y|<\delta$.

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

8. Originally Posted by Opalg
Now suppose that $\displaystyle f(x)\geqslant2\varepsilon$ for some value of x. Then (by the triangle inequality) it will follow that $\displaystyle f(y)>\varepsilon$ whenever whenever $\displaystyle |x-y|<\delta$. Therefore $\displaystyle \int_x^{x+\delta}\!\!\!f(y)\,dy>\varepsilon\delta$. But the fact that $\displaystyle \lim_{X\to\infty}\int_0^X\!\!\!f(x)\,dx$ exists tells you that there exists $\displaystyle X_0$ such that $\displaystyle \int_X^Y\!\!\!f(x)\,dx<\varepsilon\delta$ whenever $\displaystyle Y>X>X_0$ (as in the previous paragraph). Therefore $\displaystyle f(x)\geqslant2\varepsilon\:\Rightarrow\:x<X_0$. Equivalently, $\displaystyle f(x)<2\varepsilon$ whenever $\displaystyle x>X_0$. This says that $\displaystyle f(x)\to0$ as $\displaystyle x\to\infty$.
Why should we prove that $\displaystyle f(x)\geqslant2\varepsilon \implies x<X_0$? Can't we just go straight to proving $\displaystyle f(x)<2\varepsilon \implies x>X_0$ or do we need to prove the former to prove the latter?