Proving uniform convergence of $f_n(x) = \frac{x}{1 + n^2 x^2}$

Question

Let $f_n:[-1, 1] \to \mathbb{R}$ defined by $f_n(x) = \dfrac{x}{1 + n^2 x^2}$ be a sequence of functions where $n \in \mathbf{N}$.

Goal: Show that $f_n$ converges uniformly to a function by showing that $|f_n(x)| \leq \dfrac{1}{n}$.

I am able to bound it up to some point. I just couldn't show that it is precisely bounded above by $1/n$. My attempt so far:

Proof.

$f_n$ clearly converges pointwise to $0$. Now, we check uniform convergence.

If $x = 0$, then $f_n \to 0$ uniformly (because it is identically $0$) for $x \in [-1, 1]$.

Now suppose $x \neq 0$. Then,

$$|f_n(x) - 0| = \left|\dfrac{x}{1 + n^2 x^2}\right| \leq \left| \dfrac{1}{n^2 x^2}\right| \leq \left| \dfrac{1}{n^2 x}\right| = \dfrac{1}{n^2 |x|}$$

End of attempt.

Here is where I got stuck. I can see that if we can bound $f_n(x)$ above by $1/n$, then the following has to be true:

$$ \dfrac{1}{n^2 |x|} \leq \dfrac{1}{n}\iff |x| \geq \dfrac{1}{n} \iff n|x| \geq 1$$

But, I just can't see how the above is true. Am I missing something?

Answer 1

Observe that, when $\lvert x \rvert >0$, $$\lvert f_n(x)\rvert =\frac{1}{2n}M_{-1}\left(n\lvert x \rvert,\tfrac{1}{n\lvert x \rvert}\right)$$ where $M_{-1}$ denotes the harmonic mean. Let $M_0$ denote the geometric mean. the geometric mean–harmonic mean inequality states that $$M_{-1}\leq M_0\text{,}$$ with equality holding iff all operands are equal. Consequently, $$M_{-1}\left(n\lvert x \rvert,\tfrac{1}{n\lvert x \rvert}\right)\leq 1\text{,}$$ with equality holding iff $\lvert x \rvert =\tfrac{1}{n}$. Hence $$\lvert f_n(x)\rvert\leq \tfrac{1}{2n}\text{.}$$

Answer 2

Let’s look for the supremum of $f_n$. We have

$$f_n’(x)={1-n^2x^2\over (1+n^2x^2)^2}$$

So we have a minimum for $x_m=-1/n$, the minimum being $m=-1/2n$ and a maximum for $x_M=1/n$ the maximum being $M=1/2n$. So we have

$${-1\over n}\lt {-1\over 2n}\leq f_n(x)\leq {1\over 2n}\lt {1\over n}$$

Answer 3

Compute the derivative to show that $f_n(x)$ has a global maximum at $x=1/n$, and $f_n(1/n)=1/(2n)$

Answer 4

$$\frac {|x|}{1+n^2x^2}\le \frac {1}{n} \iff|nx|\le 1+n^2x^2 \iff $$

$$n^2x^2 -|nx|+1\ge 0 \iff (|nx|-1/2)^2+3/4\ge 0$$

Answer 5

$$|f_n(x)|=\frac{|x|}{1+n^2x^2}$$ $$=\frac{1}{n}\times\frac{1}{n|x|+\frac{1}{n|x|}}$$ $$\leq\frac{1}{2n}$$