Investigate whether the sequence $\frac{nx^n + 3\sin{2n\pi x}}{n}$ is uniformly convergent

Question

Let $(f_n)$ be a sequence of functions defined by \begin{equation*} f_n(x) = \frac{nx^n + 3\sin{(2n\pi x)}}{n} \end{equation*} for all $x \in [0,1]$ and $n \in \Bbb N$.

Show that $f_n$ is converges pointwise to a function $f:[0,1]\to \Bbb R$ on $[0,1]$. Investigate whether $f_n$ is uniformly convergent to $f$ on $[0,1]$ or not.

Attempt:

1. For $x=0$: We have $f_n(0)=0 \to 0$.

2. For $x=1$: We have $f_n(1)=1 \to 1$.

3. For $x \in (0,1)$: I claimed that $f_n \to 0$. The proof is goes as follows: Let $\varepsilon > 0$ and $x \in (0,1)$ be arbitrary. Choose $k \in \Bbb n$ such that $k > \max\left\{\frac{2x}{\varepsilon - \varepsilon x}, \frac{6}{\varepsilon} \right\}$. Then, for any $n \in \Bbb n$ with $n \ge k$, we have \begin{align*} \left|f_n(x)-f(x) \right| &= \left|\frac{nx^n + 3\sin{(2n\pi x)}}{n}-0 \right| \\ &\le x^n + \frac{3}{n} \\ &= \frac{1}{\left(1+\left(\frac{1}{x}-1\right)\right)^n} + \frac{3}{n} \\ &\le \frac{1}{1+n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\ &< \frac{1}{n\left(\frac{1}{x} -1 \right)} + \frac{3}{n} \\ &\le \frac{1}{k\left(\frac{1}{x} -1 \right)} + \frac{3}{k} \\ &< \frac{\frac{\varepsilon}{2}\left(\frac{1}{x} -1 \right)}{\frac{1}{x}-1} + \frac{\varepsilon}{2} \\ &= \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Hence, $f_n \to f$ where \begin{equation*} f(x) = \begin{cases} 0, \qquad \text{if} \ 0 \le x < 1 \\ 1, \qquad \text{if} \ x = 1 \end{cases}. \end{equation*} Now, my intuition is said that $f_n$ is not uniformly convergent to $f$ on $[0,1]$. But, I didn't be able yet to find the subsequence such that the sequence did not uniformly convergent. On the other hand, I tried using the uniform norm \begin{align*} ||f_n - f||_{[0,1]} = \sup \left\{\begin{array}{lr} \frac{nx^n + 3\sin{(2n\pi x)}}{n}, & \text{for} \ 0 \le x < 1\\ 0, & \text{for} \ x=1 \end{array}\right\} = \frac{13}{4} \ne 0. \end{align*} But, is this correct? Any ideas? Thanks in advanced.

Answer 1

If $x\in[0,1)$, then$$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}x^n+\frac{3\sin(2\pi x)}n=0$$and you always have $f_n(1)=1$. Therefore, your sequence converges pointwise to$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<1\\1&\text{ if }x=1,\end{cases}\end{array}$$which is a discontinuous function. Since each $f_n$ is continuous, the convergence is not uniform.

Answer 2

There is simple way of showing that the convergence is not uniform. Note that $f_n(x)=x^{n}+g_n(x)$ where $g_n(x)=\frac {3\sin (2\pi nx)} n$. Now $g_n(x) \to 0$ uniformly so $f_n$ converges uniformly if and only if $x^{n}$ does. Can you show that $x^{n}$ does not converge uniformly?