Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've got another question relating to the uniform convergence of a sequence of functions (sorry). The sequence of functions in question is:

$f_n(x) = \displaystyle\frac{x}{1+n^2x^2}$ on the interval $[0,1]$

I've already shown that $h_n(x) = \displaystyle\frac{nx}{1+n^2x^2}$ converges, but not uniformly by looking at $x = \frac{1}{n}$

But for $f_n(x)$ clearly $\displaystyle\lim_{x \to +\infty} f_n(x) = f(x) = 0$ for every $x \in [0,1]$

So if $f_n(x)$ were uniformly convergent we would have $\bigg|\displaystyle\frac{x}{1+n^2x^2}\bigg| = \bigg|\frac{1}{\frac{1}{x} + n^2x}\bigg| < \epsilon \ $ for some given $\epsilon > 0$, $n > N \in \mathbb{N}$ and $\forall x \in [0,1]$

Or, alternatively, by Cauchy's criterion we would have

$|f_n(x) - f_m(x)| \leq \bigg|\displaystyle\frac{1}{n^2x} - \frac{1}{m^2x}\bigg| < \epsilon$

For $m > n > N$ and $\forall x \in [0,1]$

The problem is, I'm not convinced if $f_n(x)$ is uniform or not, because while, given any $\epsilon$, we could find a smaller $x$ to increase the value of $\bigg|\frac{1}{n^2x}\bigg|$ this is counteracted by the $\frac{1}{x}$ term also on the denominator. So I'm not sure where to go in regards to proving whether this function is uniformly convergent or not.

Thanks in advance!

share|improve this question
1  
Can you find the maximum value of $f_n$ on $[0,1]$? The derivative is not hard to analyze. –  David Mitra Feb 26 '13 at 12:18
    
@David Hi David, this question is asked prior to covering derivatives in analysis so I wanted to answer it without using differentiation. Thanks for the suggestion though! –  Noble. Feb 26 '13 at 12:19

2 Answers 2

up vote 4 down vote accepted

Basically, there are two useful observations here, valid for all $x$:

$$f_n(x) < x \quad \text{and} \quad f_n(x) < \dfrac1{1+n^2x^2}$$

Let $\epsilon > 0$ be arbitrary. If $x < \epsilon$ then by the first estimate, $|f_n(x) - f(x)| = f_n(x) < \epsilon$. Next we observe that the second estimate $\dfrac1{1+n^2x^2}$ is monotonically decreasing for $x \in [0,1]$.

It is thus sufficient (together with our earlier observation) to pick an $N$ satisfying $\dfrac1{1+N^2\epsilon^2} < \epsilon$; the monotonicity then provides $\dfrac1{1+N^2x^2} < \epsilon$ for all $x \in [\epsilon, 1]$.

Hence $f_n$ is uniformly convergent on $[0,1]$. The method used is quite common: to use a different estimate close to one of the boundaries, thereby obtaining a new interval of uniform convergence that permits an easier assessment (for example because the bounds of the interval depend on $\epsilon$, as in this particular case).

share|improve this answer
    
Thanks a lot for the helpful explanation. –  Noble. Feb 26 '13 at 12:18
    
You're welcome, HTH. –  Lord_Farin Feb 26 '13 at 12:35

Fix $n$ and study the function $f_n$ on $[0,1]$.

We have $$ f_n'(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}. $$

So $f_n$ is nonnegative and attains its maximum at $1/n$ on $[0,1]$.

Hence $$ 0\leq \sup_{[0,1]} f_n=f_n(1/n)=\frac{1}{n}. $$

By the squeeze theorem, it follows that $f_n$ converges uniformly to $f(x)=0$ on $[0,1]$.

share|improve this answer
    
A small correction. $f_n(\frac{1}{n}) = \frac{1}{2n}$. –  user154185 Jun 7 at 16:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.