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