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The question is:

Define $f_k: [0,1] \rightarrow \mathbb{R}$ by $$ f_{k}(x) = \begin{cases} 2kx^2 & 0 \leq x \leq \frac{1}{2k} \\ 2k^2(\frac{1}{k}-x) & \frac{1}{2k} \leq x \leq \frac{1}{k} \\ 0 & \frac{1}{k} \leq x \leq 1 \end{cases}$$ Prove that $(f_{k})_{k \in \mathbb{N}}$ converges pointwise to zero on $[0,1]$.

The definition I have is as follows

A sequence $(f_k)_{k \in \mathbb{N}}$ of functions $f_k:A \rightarrow \mathbb{R}$ converges pointwise to a function $f:A \rightarrow \mathbb{R}$ if $\forall \epsilon > 0, \forall x \in A, \exists N \in \mathbb{N}$ such that $\forall k>N: |f_{k}(x)-f(x)|<\epsilon$

How do I go about proving this via the definition?

My attempt is currently

Fix $\epsilon>0$. If $x=0$ then $f_{k}(0)=0$. Now for $0 \leq x \leq \frac{1}{2k}$ we have that $|f_{k}(x)-f(x)|=|2k^{2}x| \leq |k|$ on the interval since $x\in[0,\frac{1}{2k}]$.

I can also find a similar inequality for the second case, but I don't understand how to find a suitable $N$.

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    You don't need to *find* a suitable $N$, just show that one exists. For example, given any $x \in (0,1]$, there exists an $N$ such that $\frac{1}{N} < x$. – 2017-01-09
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    @MorganRodgers I don't think I follow. To show one exists we must first find one right? Like in an epsilon-delta continuity proof where we choose a $\delta$? – 2017-01-09
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    No, you don't need to find one if you have another way to show it exists. The fact that there exists an $N$ with the property I mentioned is a standard property of the real numbers. This goes with the answer suggested by Jorge; there is an $N$ such that $N > 1/x$ for any $x$. I know this because I know that the natural numbers are not bounded. I don't need to know what $x$ is, and I don't need to give a formula for $N$. – 2017-01-09
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    Also in $\epsilon$-$\delta$ proof, you don't need to explicitly give a $\delta$; you only need to show that there exists a $\delta$. – 2017-01-09
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    Right thanks I understand. For my purposes in my exams I believe my lecturer prefers me to explicitly write down an $N$ (and in $\epsilon-\delta$ proofs to give a $\delta$). – 2017-01-09

1 Answers 1

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What we want to prove is that those functions converge pointwise to the constant zero function.

Take $\epsilon > 0$ and $x\in [0,1]$. If $x\neq 0$ then take $N$ such that $N>\frac{1}{x}$, notice that if $m\geq N$ then $k > \frac{1}{x}$ and so $x>\frac{1}{k}$ meaning $f_k(x)=0$ and so $|f_k(x)-0|<\epsilon|$.

If $x=0$ then wwe always have $f_k(x)=2kx^2=0$, so that case is also done.

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    What made you choose $N>\frac{1}{x}$? In general how would you find $N$? – 2017-01-09