Determining Pointwise Convergence of Sequences of Functions

Recall from the Pointwise Convergence of Sequences of Functions page that a sequence of functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be pointwise convergent if for all $x \in X$ and for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

(1)

\begin{align} \quad \mid f_n(x) - f(x) \mid < \epsilon \end{align}

We will now look at some examples of determining whether a sequence of functions is pointwise convergent or divergent.

For example, consider the following sequence of functions with common domain $\mathbb{R}$:

(2)

\begin{align} \quad (f_n(x))_{n=1}^{\infty} = \left ( \frac{\sin (2x + 3n)}{n^2} \right )_{n=1}^{\infty} = \left ( \sin (2x + 3), \frac{\sin (2x + 6)}{4}, \frac{\sin (2x + 9)}{9}, ... \right ) \end{align}

We claim that this sequence converges pointwise to the limit function $f(x) = 0$. To show this, we note that for all $x \in X$ and for all $n \in \mathbb{N}$ that:

(3)

\begin{align} \quad -1 \leq \sin (2x + 3n) \leq 1 \end{align}

Therefore we see that:

(4)

\begin{align} \quad -\frac{1}{n^2} \leq \frac{\sin (2x + 3n)}{n^2} \leq \frac{1}{n^2} \end{align}

Now take the limit as $n$ goes to $\infty$ from both sides:

(5)

\begin{align} \quad \lim_{n \to \infty} -\frac{1}{n^2} \leq \lim_{n \to \infty} \frac{\sin (2x + 3n)}{n^2} \leq \lim_{n \to \infty} \frac{1}{n^2} \end{align}

Since $\lim_{n \to \infty} - \frac{1}{n^2} = 0$ and $\lim_{n \to \infty} \frac{1}{n^2} = 0$ we have by the Squeeze theorem that then $\lim_{n \to \infty} \frac{\sin (2x + 3n)}{n^2} = f(x)$ where $f(x) = 0$, and so the sequence $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x) = 0$.

For another example, consider the following sequence of functions with common domain $[0, 1]$:

(6)

\begin{align} \quad (g_n(x))_{n=1}^{\infty} = \left ( \frac{x^2 + 2x + x}{n^2 + 2n + n} \right )_{n=1}^{\infty} \end{align}

Note that since $0 \leq x \leq 1$ that then $0 \leq x^2 \leq 1$ and $0 \leq 2x \leq 2$, and so for $x \in [0, 1]$ we see that

(7)

\begin{align} \quad 0 \leq x^2 + 2x + x \leq 4 \end{align}

Therefore we have that:

(8)

\begin{align} \quad 0 \leq \frac{x^2 + 2x + x}{n^2 + 2n + n} \leq \frac{4}{n^2 + 2n + n} \end{align}

Taking the limit from both sides as $n \to \infty$ and we get that:

(9)

\begin{align} \quad 0 \leq \lim_{n \to \infty} \frac{x^2 + 2x + x}{n^2 + 2n + n} \leq \lim_{n \to \infty} \frac{4}{n^2 + 2n + n} = 0 \end{align}

Once again by the squeeze theorem we have that $(g_n(x))_{n=1}^{\infty}$ converges pointwise to the limit function $g(x) = 0$.