Pointwise convergence of a sectionally defined function

Question

I'm asking for a proof-verification of the following:

$ f_n(x) := \left\{\begin{array}{ll} 2n^2x, & x\in [0,\frac{1}{2n}) \\ -2n^2x+2n, & x\in [\frac{1}{2n},\frac{1}{n})\\ 0,& x \in [\frac{1}{n},1]\end{array}\right. . $

Claim: $f_n$ converges pointwise to $f$ where $f(x):= 0$.

Proof:

Let $\epsilon_1 := 2(n+1)^2x \,\, and \,\, x \in [0,1]$

$\Rightarrow \mid f_n(x) - f(x) \mid = \mid f_n(x) - 0\mid = \mid f_n(x) \mid$

Case one: $x \in [0,\frac{1}{2n})$

$\mid 2n^2x \mid = 2n^2x < \epsilon_1 \forall n$

Case two: $x \in [\frac{1}{2n}, \frac{1}{n})$ Let $\epsilon_2 > 0,$ $\epsilon_2 := 2(n+1)^2x+2(n+1)$

$\mid -2n^2x+2n\mid \leq \mid -2n^2x \mid + \mid 2n \mid = 2n^2x +2n < \epsilon_2 \forall n$

Case three: $ x \in [\frac{1}{n},1]$ Let $ \epsilon_3 > 0, \epsilon_3 := 1$

$\mid 0 \mid = 0 < \epsilon_3 \forall n$

$\Rightarrow f_n$ converges to $f$ when $n \rightarrow \infty $ where $f(x) := 0$

Answer 1

Since your $\varepsilon$'s depend upon $n$ and $x$, your proof is wrong. There can be no restriction on the value of $\varepsilon$, apart from the fact that we must have $\varepsilon>0$.

There are two cases to be considered:

• $x=0$. Then each $f_n(0)$ is equal to $0$ and therefore $\lim_{n\to\infty}f_n(0)=0$.
• $x\in(0,1]$: then take $N\in\mathbb N$ such that $\frac1N\leqslant x$. Then $n\geqslant N\implies f_n(x)=0$. Therefore, again, $\lim_{n\to\infty}f_n(x)=0$.

EDIT:

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