$f_n : [0,1] \rightarrow \mathbb{R}$ for each $n \in \mathbb{N}$

$ f_n(x) = \left\{ \begin{array}{lr} 1 - nx & : 0 \le x \le \frac{1}{n}\\ 0 & : \frac{1}{n} \le x \le 1 \end{array} \right. $

Show the following:

(a) $\{f_n\}$ converges pointwise on $[0,1]$. What is the limit function?

(b) $\{f_n\}$ doesn't converge uniformly on [0,1]

(c) if $0<\epsilon<1$ then $\{f_n\}$ converges unifomrly on $[\epsilon,1]$

(a) for $0 \le x \le \frac{1}{n}$ , Does $\lim_{n\to\infty} (1-nx) = 1$ or $=-\infty$ ?

If $= -\infty$, Would the limit function be $f_n(x) = -n$?

I feel like $\lim_{n\to\infty} (1-nx) = 1$ because as $n \rightarrow \infty$ our upper bound $\frac{1}{n} \rightarrow 0$. Therefore, $\{f_n\}$ converges pointwise on [0,1]. If this is the case, would our limit function simply be $f_n(x) = 1 - 0n = 1$?

(b) I believe our $\{f_n\}$ is discontinuous at $0$, which causes the problem, but do we need to find the $ \lim_{n\to\infty} \sup f_n(x) \nrightarrow 0$ ? I am a little confused about this part.

(c) Since $0$ is where $\{f_n\}$ has issues, if we take a point, call it $\epsilon$, s.t. $0< \epsilon<1$, are we trying to show discontinuity at $0$ is what makes the function not converge uniformly?

Any help is greatly appreciated!



(a) Let $x > 0$. What is $f_n(x)$ when $n > \frac{1}{x}$? Does that help you find $\lim_{n \to \infty} f_n(x)$?

What is $\lim_{n \to \infty} f_n(0)$?

(b) What is the definition of uniform continuity? Denote by $f$ the pointwise limit of $f_n$. Is it possible to find a universal $N$ such that, for each $x$, $|f_n(x) - f(x)| < \frac12$ whenever $n \geq N$? Think about points that are very close to $0$.

(c) Let $\epsilon > 0$. If $n > \frac{1}{\epsilon}$, what are the values of $f_n$ in the interval $[\epsilon, 1]$?

Hope this gets you started.

  • $\begingroup$ Thanks! I appreciate the hints! Helped me figure this out. $\endgroup$ – user161154 Feb 13 '15 at 16:52

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