Uniform Convergence of $\langle f_n \rangle = (nx)/(1+n^3x^2)$ for $0 \leq x \leq 1$

Question

Uniform Convergence of $\langle f_n \rangle = (nx)/(1+n^3x^2)$ for $0 \leq x \leq 1$

I was to prove the above statement. Instead I somehow did something wrong and proved that it isn't uniformly convergent. Please check where I went wrong.

$$ \langle f_n\rangle= (nx)/(1+n^3x^2)\\ \lim_{n\to\infty} f_n(x)= f(x)=0 $$ We choose $\varepsilon = \frac{1}{4}$. $$|f_n(x)-f(x)|\lt\varepsilon$$ Since $x \in [0,1]$, we choose $x=1/2$ and $n=2$. Then $$ 1/3 \not\lt 1/4 $$

Answer 1

The inegality you wrote does not prove anything. The negation of the definition of the uniform convergence is :

$$\exists \epsilon > 0, \forall A \in \mathbb{N}, \exists N_\epsilon\ge A / (\exists x \in [0,1], ||f_{N_\epsilon}(x)-f(x)||>\epsilon)$$

Here is one way to prove the uniform convergence :

• Where does $f_n$ reach its maximum ?
• What is the maximum at this point ?
• To what converge the maximum when $n \to \infty$ ?
• What can you conclude for $f_n$ ?

Answer 2

Late to the game, but: we want to show that $f_{n} \to f$ uniformly, where $f(x) = 0$ for $x \in [0, 1]$. Note $|f_{n}(x) - f(x)| = \frac{x}{1 + n^{3}x^{2}} = f_{n}(x)$; we find the maximum of this function. $f_{n}'(x) = \frac{1 - n^{3}x^{2}}{(1 + n^{3}x^{2})^{2}}$ and $f_{n}'(x) = 0 \Rightarrow x^{*} = \frac{1}{n^{3/2}}$; note $0 \leq x^{*} \leq 1$ for all $n \in \mathbb{N}$. Thus, $|f_{n}(x) - f(x)| \leq \frac{1/n^{3/2}}{1 + n^{3}(1/n^{3})} = \frac{1}{2n^{3/2}}$ for $x \in [0, 1]$, $n \in \mathbb{N}$. Now, fix $\epsilon > 0$. Take $N = N(\epsilon) = \lfloor\frac{1}{(2\epsilon)^{2/3}} \rfloor + 1 $ to see that for $n \geq N$, we have $n \geq N > \frac{1}{(2\epsilon)^{2/3}} \Rightarrow \frac{1}{2n^{3/2}} < \epsilon \Rightarrow |f_{n}(x) - f(x)| < \epsilon$. This proves uniform convergence.