prove $({f_n})_n$ is uniformly convergent on ${[0,1]}$

Question

The real function ${g}$ is continuous on $[0,1]$ .we define ${f_n}$ on ${[0,1]}$:

$$f_n(x)=\frac{{{g(x)\sin^{n} (x)}}}{{{1+nx}}}$$

prove $({f_n})_n$ is uniformly convergent on ${[0,1]}$ .

Answer 1

Since $g$ is continuous on $[0,1]$, it is bounded by some number $K$. Given $\varepsilon > 0$, let $N$ be a positive integer greater than or equal to $\frac{K}{\varepsilon}$. Let $n \ge N$ and $x\in [0,1]$. By the mean value theorem, $|\sin x| \le x$, hence $|\sin^n x| \le x^n$. Therefore

$$|f_n(x)| \le \frac{|g(x)|x^n}{1 + nx} \le K\frac{x^n}{1 + nx}.$$

For each $n \in N$, the function $t\mapsto \frac{t^n}{1 + nt}$ is increasing on $[0,1]$ since

$$\frac{d}{dt}\left(\frac{t^n}{1 + nt}\right) = \frac{nt^{n-1}(1 + nt) - nt^n}{(1 + nt)^2} = \frac{nt^{n-1}(1 + (n-1)t)}{(1 + nt)^2} \ge 0.$$

Therefore $\frac{x^n}{1 + nx} \le \frac{1}{1 + n}$, and thus $$|f_n(x)| \le \frac{K}{1 + n} < \varepsilon.$$ Since this inequality holds for every $x\in [0,1]$ and every $n \ge N$, $\sup_{x\in [0,1]} |f_n(x)| < \varepsilon$ for all $n\ge N$. Consequently, $f_n$ converges to $0$ uniformly on $[0,1]$.