I would like some help in checking the following epsilon-delta proof I have written on a case-by-case basis for the pointwise convergence of the following function.
I am trying to prove that the sequence of functions $g_n(x)$ tends to $f(x) = 0$ in a pointwise fashion,
where $g_n(x) = \begin{cases} 2nx & \text{for } x \in [0, \frac{1}{2n}) \\ -2n(x - \frac{1}{n}) & \text{for } x \in [\frac{1}{2n}, \frac{1}{n}) \\ 0 & \text{for } x \in [\frac{1}{n}, 1] \end{cases}$
I do understand and have internalised from reading other explanations (but I have not seen any case-by-case epsilon delta proofs for this online, which is what I am really trying to create myself as it would help my understanding) that for any point I pick $x$, the triangle of the graph, i.e. all non-zero values eventually pass to the left of it, so eventually $f_n(x)$ as I keep increasing $n$ and have a fixed $x$, reaches a point after which $f_n(x) = 0$, so $f_n$ converges pointwise for all nonzero $x$ (as we fix $x$ and then would find the right $N$ for it, i.e. when $x > \frac{1}{n}$ so $n > \frac{1}{x} = N$). But what pitfalls have I walked into when I try the following approach to showing pointwise convergence to the zero function?
My attempy for a case-by-case proof:
To show pointwise convergence, it means that I want to show that for any $x \in [0, 1]$, for any $\epsilon > 0$ I pick, there exists a natural number $N$ such that for a particular $x$, $|g_n(x) - 0| < \epsilon$.
For the case $x = 0$: $g_n(x) = 0$ for all $n$ in natural numbers (so $N = 1$)
For the case $x \in [\frac{1}{2n}, \frac{1}{n})$: I have $|-2n(x - \frac{1}{n})-0| \leq 2n |x - \frac{1}{n}|$
Since $x$ is at most 1, $\frac{1}{2}, \frac{1}{3}, \ldots$ and $\frac{1}{n}$ is 1, $\frac{1}{2}, \frac{1}{3}, \ldots$
$2n |x - \frac{1}{n}| = 2n \left(\frac{1}{n} - x\right) = 2n | \frac{1}{n} - x | $, then for any $n$ greater than some natural number $N$,
$2n | \frac{1}{n} - x | = |2 - 2nx| \leq |2 - 2Nx| \leq |2| + |2Nx| = \epsilon$, then $N = \frac{\epsilon - 2}{2x}$ for $x \in [\frac{1}{2n}, \frac{1}{n}]$
For the case $x \in [0, \frac{1}{2n})$: I want to find an $N$, related to the $\epsilon$ I pick, such that for any $n$ greater than $N$, $|2nx - 0| < \epsilon$, so I try to bound $|2nx|, |2nx| \leq |2n(\frac{1}{2n})|$ as $x$ is at most $\frac{1}{2n}$, more specifically at most $\frac{1}{2}$, then $|2n(\frac{1}{2n})| \leq |2n(\frac{1}{2})| = |n|$
For the case $x \in [\frac{1}{n}, 1)$: $g_n(x) = 0$, so $N = 1$ as for any $x$, given any $\epsilon$, taking $N = 1$ means that $|g_n(x) - 0|$ will be less than $\epsilon$ for all $n \geq N$.
Your argument seems like overkill.
If $x > 0$, then there exists $N \in \mathbb{N}$ such that $1/N < x$ (this follows directly from the Archimedean property, though you can certainly find a precise bound for $N$ with a little bit of basic algebra). If $n > N$, then $g_n(x) = 0$. Hence $g_n(x) \to 0$ for any $x > 0$.
On the other hand, $g_n(0) = 0$ for all $n$, and so $g_n(0) \to 0$.
Hence $g_n$ converges pointwise to zero.