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I have the following problem. For each positive integer $n$, let $f_n: [0, 2] \rightarrow \mathbb{R}$ defined by $$f_n(x) = \begin{cases} nx & \text{if $0 \le x \le\frac{1}{n}$} \\ 1 & \text{if $\frac{1}{n} < x \le 2$} \end{cases}$$ Let $f: [0, 2] \rightarrow \mathbb{R}$ defined by $f(x) = 0$ for all $x \in (0, 2]$. Prove that $\langle f_n \rangle_{n = 1}^\infty$ does not converge uniformly to $f$ by negating the definition of uniform convergence. I guess the negation is that there exists an $\epsilon > 0$, an $x \in [0, 2]$ and an positive integer $n$ such that $\lVert f_n - f \rVert = \sup \Big \{ \lvert f_n (x) - f(x) \rvert: x \in [0, 2] \Big \} \ge \epsilon$. I am not sure if I am negating correctly.

At the same time, I notice that if $\frac{1}{n} < x \le 2$, $\langle f_n \rangle_{n = 1}^\infty$ does not converge pointwise to $f$, so it cannot converge uniformly to $f$. I am a little bit stuck when $0 ] \le x \le \frac{1}{n}$. I am not sure if I am doing things right because it seems like I am not using the negation at all. Please help me if you can. Thank you so much.

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2 Answers 2

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Hint:
Given any $x\in (0,2]$, there exists $N$ such that $\frac 1n\lt x$ for all $n\ge N$. It follows that for $n\ge N, f_n(x)=1$ and so $f_n(x)\to 1$ but $f_n(0)\to 0$.

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Let $n$ be any positive integer. $\|f_n-f\|\geq |f_n(\frac 1 n) -f(\frac 1 n)|$ since the supremeum of a set is greater than or equal to each element of the set. Hence $\|f_n-f\|\geq |1-0|=1$ which shows that $f_n$ does not converge to $f$ uniformly.

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