Define a sequence of functions, $g_n$ on [0,1] and explain why they converge point wise on [0,1]

Question

\begin{equation*} g(x)=\begin{cases} n^2x \quad &\text{for} \, x \in\ [0,1/n] \\ 2n-n^2x \quad\ &\text{for} \, x \in\ [1/n,2/n] \\ 0 \quad &\text{for} \, x \in [2/n,1]\\ \end{cases} \end{equation*}

I know the sequence converges to basically being near infinity at the origin and zero everywhere else, but how would I prove that it converges point wise on [0,1], especially if it seems to approach infinity at one point?

Also, the sequence is technically increasing on the first interval, decreasing on the second, neither on the third, but neither increasing nor decreasing on the entire interval as a whole, correct?

Answer 1

Note that $g_n (0) = 0$ for any n. Given any fixed $x > 0$, there exists $N$ large enough so that $x > \frac{2}{n}$ for all $n > N$ (choose $N>\frac{2}{x}$, for instance). Thus, for any fixed $x > 0$, we have $g_n (x) \equiv 0$ for large enough $n$, so $g_n (x) \to 0$ for each positive $x$. Since $g_n(0)$ is always zero, $g_n(x) \to 0$ at each point $x \in [0,1]$.

I would say it doesn't make much sense so say "the sequence is increasing on the first interval", since the definition of "the first interval" changes with $n$.

This is a very important example, which illustrates that pointwise convergence is not the be-all end-all measure of convergence of functions (and this fact is central to modern analysis).

Answer 2

For $x>0$, it will eventually belong to the interval $[2/n,1]$ when $n$ is sufficiently large. Therefore $g_n(x)=0$ in this case. When $x=0$, $g_n(x)=0$ for all $n$, and thus the limit is also $0$.