# How to prove a sequence of a function converges uniformly?

For $$n \in \mathbb{N}$$, define the formula, $$f_n(x)= \frac{x}{2n^2x^2+8},\quad x \in [0,1].$$ Prove that the sequence $$f_n$$ converges uniformly on $$[0,1]$$, as $$n \to \infty$$.

I know that the definition says $$f_n$$ converges uniformly to $$f$$ if given $$\forall \epsilon \gt 0$$, $$\forall n \geq N$$, such that $$|f_n(x) - f(x)| \lt \epsilon, \forall n \geq N$$ and $$\forall x \in [0,1].$$

I looked first at the pointwise convergence and found that $$\lim_{n \rightarrow \infty} \frac{x}{2n^2x^2+8} = 0, \forall x \in [0,1].$$

So how do I use this to choose an $$n \geq N$$ such that $$|f_n(x) - f(x)| \lt \epsilon$$ ?

Right now, I have

"proof: Let $$\epsilon > 0, \exists N \in \mathbb{N}$$ such that, n $$\geq N \Rightarrow \frac{1}{2n^2+8} \lt \epsilon$$,

by $$|f_n(x) - 0| = |\frac{x}{2n^2x^2+8}| \leq |\frac{x^2}{2n^2x^2+8}| \leq \frac{1}{2n^2x^2+8} \;\;\;\; \forall x \in [0,1].$$

Since $$\lim_{n \rightarrow \infty} \frac{x}{2n^2x^2+8} = 0, \forall x \in [0,1]$$, $$f_n(x)$$ will converge uniformly to $$0$$ on $$[0,1]$$."

Is this correct? Am I missing something? Is something not correct? I'm unsure about my choice of $$N$$. Please & thanks!

• Hi Jamie, you seemed to have proved the sequence converges pointwise, but not uniformly. – Suugaku Apr 23 '13 at 4:37
• What is missing from my proof to make it uniform? I thought if I proved it pointwise, and then showed that it converges $\forall n \geq N$ and $\forall x \in [0,1]$, then that implies uniform convergence? – Jaime Apr 23 '13 at 5:24

related problem: (I), (II), (III). Here is a systematic technique. In order to find $\sup_{0\leq x\leq 1} |f_n(x)-f(x)|$, you need to maximize the function $\Big|\frac{x}{2n^2x^2+8}\Big|$ over the interval $[0,1]$. Now, let

$$g(x)=\frac{x}{2n^2x^2+8} \implies g'(x) = \frac{4-n^2 x^2}{(2n^2x^2+8)^2}=0 \implies x=\frac{2}{n}$$

gives the max of the function $g(x)$ which is $g(2/n)=1/8n$. You can check this by checking the sign of $g''(x)$ which should be $< 0$. Hence we have

$$\sup_{0\leq x\leq 1} |f_n(x)-f(x)|= \sup_{0\leq x\leq 1} \Big|\frac{x}{2n^2x^2+8}\Big|=\frac{1}{8n} < \epsilon.$$

• Using your technique, which I have seen in three posts already, how would you apply it to a sequence of functions of this form: $f_n(x)=x^{1+\frac{1}{2n-1}}$ – funmath May 5 '18 at 19:06

You can write $$f_n(x)={1\over n} g(n\>x)\qquad(0\leq x\leq 1)$$ with $$g(t)={t\over 2t^2+8}\qquad(0\leq t<\infty)\ .$$ Since $g(t)$ converges to $0$ for $t\to\infty$ we already can say that $g$ is bounded. A quantitative estimate can be obtained as follows: For $t>0$ one has $$0<g(t)={1\over 8}\ {2\over{t\over 2}+{2\over t}}\leq{1\over 8}\ .$$ Therefore we now have $$|f_n(x)|\leq{1\over 8n}\qquad(n\geq1,\ 0\leq x\leq1)\ ,$$ which shows that the $f_n$ converge uniformly to $0$ on $[0,1]$.

Another way to maximize :

$\dfrac{x}{2n^2x^2+8}\leq \dfrac{x}{8nx}=\dfrac{1}{8n}$ where we used the AM-GM inequality :http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means to show

$\dfrac{2n^2x^2+8}{2} \geq \sqrt {2n^2x^2 \cdot 8} \ \ \ \$ Applying this easy inequality is often a good first way to find a supremum for the function if you have a summation involved. So since the function converges pointwise to $f(x)=0$ We have the following result:

$\lim_{n \to \infty}\sup_{0\leq x\leq 1} |f_n(x)-f(x)|=\lim_{n \to \infty}|\dfrac{1}{8n}|=0$

Pointwise convergence is not enough to say that the function converges uniformly. Here, $$f_n(x)= \frac{x}{2n^2x^2+8},x \in [0,1]$$ has pointwise convergence to $f(x)=0$, so by definition $$|f_n(x)-f(x)|=\bigg|\frac{x}{2n^2x^2+8}-0\bigg|= \bigg|\frac{x}{2n^2x^2+8}\bigg|< \frac{1}{2n}$$ This shows that that the function is not uniformly convergent.

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jan 19 '16 at 17:00