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Let $h_n: [a,1] \to R$ defined by:

$h_n(x) = n^2x$ for $a \le x < \frac 1n$ and $h_n(x) = \frac 1x$ for $ \frac 1n \le x \le 1$.

Show that $\{h_n\}_{n=1}^\infty$ converges uniformly on $[a,1]$ where $0 < a < 1$.

I guess my main concern here is what function $h(x)$ does $\{h_n\}_{n=1}^\infty$ converge pointwise to? I think its $h(x) = \frac 1x$ but I'm not sure. After I figure that out I'm pretty sure I can satisfy the necessary epsilon inequality for uniform convergence.

Thanks for the help!

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    $\begingroup$ What is $a$? For each $x>0$ there is an $m\in\Bbb Z^+$ such that $\frac1n\le x$ whenever $n\ge m$. What does this tell you about $\lim_nf_n(x)$ for $x>0$? Then, as a separate calculation, what is $\lim_nf_n(0)$? If you can answer those, you can answer your own question for any $a\in[0,1]$. $\endgroup$ – Brian M. Scott Dec 2 '14 at 7:45
  • $\begingroup$ Don't the value of those limits depend on where x is in the domain because it is a piece-wise function? $\endgroup$ – jlang Dec 2 '14 at 7:48
  • $\begingroup$ For $x\in[0,1]$ the only relevant distinction is between $x=0$ and $x>0$. $\endgroup$ – Brian M. Scott Dec 2 '14 at 7:49
  • $\begingroup$ $0$ isn't in the domain of this function though? $a > 0$. I know this function doesn't converge uniformly on $[0,1]$ $\endgroup$ – jlang Dec 2 '14 at 7:51
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    $\begingroup$ Yes. Every $h_n$ for sufficiently large $n$ is $\frac1x$ on $[a,1]$. $\endgroup$ – Brian M. Scott Dec 2 '14 at 7:57
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Take $n_0>1/a$. For all $n\ge n_0$, $x\in[a,1]$: $1/n<1/n_0<a\le x$, i.e, you are in the second case and $f_n(x)=1/x$. And uniform convergence $\implies$ pointwise convergence to tha same limit.

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Your guess is correct: the sequence of functions converges pointwise to $h(x) = 1/x$ on any interval $[a, 1]$ with $0 < a < 1$. This is because $1/n$ eventually gets smaller than $a$, so for this $n$ (and all subsequent values of $n$), $h_n(x) = 1/x$ because $1/n < a \le x \le 1$.

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  • $\begingroup$ I'm assuming you have to split this up into two cases depending on where x is in the domain, but does this look right for the first inequality? For $N > \frac 1\epsilon$ and $n \ge N$ $$|nx^2- \frac 1x| \le |n(\frac 1n)^2 - n| = | \frac {1-n^2}n| < |\frac 1n| < \frac 1N < \epsilon $$ $\endgroup$ – jlang Dec 2 '14 at 8:05
  • $\begingroup$ $\big|\frac{1-n^2}{n}\big|$ is certainly not less than $\frac{1}{n}$! $\endgroup$ – Unit Dec 2 '14 at 8:12
  • $\begingroup$ Oh wow you're totally right, what a stupid mistake. Is the first part correct though? $\endgroup$ – jlang Dec 2 '14 at 8:16
  • $\begingroup$ can you say? $$|\frac {1-n^2}n| < |\frac{n^2}n|$$ since n is a positive integer. $\endgroup$ – jlang Dec 2 '14 at 8:23
  • $\begingroup$ You do not have to deal with two cases separately. $n^2x$ and $1/x$ look nothing like each other, so this is a relief! Let $\epsilon > 0$. Pick $N > 1/a$. Then for any $x$ in $[a, 1]$ and for any $n \ge N$, we have the very easy inequality $|h_n(x) - h(x)| = 0 < \epsilon$. $\endgroup$ – Unit Dec 2 '14 at 8:30

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