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I don't understand how this piecewise converges to $0$

Determine the point wise limit of $(f_n)$ on the indicated interval, and decide whether $(f_n)$ converges uniformly to this function

$f_n = \left\{\begin{matrix} 0 & x \leq n \\ x-n & x \geq n \end{matrix}\right.$ on $[a,b]$, and on $\mathbb{R}$

So since $n \geq 0$, I only need to look at $x \geq 0$. For $[a,b]$, as $n\to\infty$, $f_n$ doesn't seem to have a pointwise limit and on $\mathbb{R}$, it also doesn't seem to have a limit as it becomes "periodic"

The answer book says it converges to $0$ and it says it doesn't converge uniformly on $\mathbb{R}$, but it does converge uniformly on $[a,b]$. Why?

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3 Answers 3

up vote 3 down vote accepted

Fix a point $x \in \mathbb{R}$ (or $[a,b]$). Then there is some $n_0 \in \mathbb{N}$ with $n_0 > x$, and therefore $f_n(x) = 0$ for all $n \ge n_0$. It should thus be clear why it converges uniformly on $[a,b]$ $-$ the sequence is eventually constant on any finite interval!

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Well I graphed the sequence of functions on an closed interval [a,b] and it terminates to a slanted line. Making more $n$s doesn't matter since they won't even show up on [a,b] –  sidht Jan 6 '13 at 21:50
    
@sizz: I'm not sure what you're saying. The point is that if you take any finite interval $[a,b]$ then for any $n>b$ all the functions will be constant (with value zero) on that interval... which is why $(f_n)$ converges on any given finite interval $[a,b]$. It's true that the slanted line eventually appears; indeed, that's why it doesn't converge uniformly on $\mathbb{R}$. –  Clive Newstead Jan 6 '13 at 21:53
    
Let $a = 3$ and $b = 4$. Then for small $n$ (say $n = 2$), $f_n$ is a "slanted line" (in this case, its graph is a line segment from $(3, 1)$ to $(4, 2)$), but for large $n$ (say $n = 7$), $f_n$ is zero on $[3, 4]$. –  Tanner Swett Jan 6 '13 at 21:54
    
@TannerL.Swett, how can we say for $n=7$ it is $0$? It doesn't even appear on $[3,4]$? –  sidht Jan 6 '13 at 22:03
1  
What do you mean, it doesn't appear? The definition states that if $x \le n$, then $f_n(x) = 0$. We are assuming that $n = 7$. For all $x \in [3, 4]$, $x \le 7$. Therefore, for all $x \in [3, 4]$, $f_7(x) = 0$. –  Tanner Swett Jan 6 '13 at 22:10

Fix $x\in\mathbb{R}$. For all $n$ large enough ($n\geq \lfloor x\rfloor +1$), we have $f_n(x)=0$. So $\lim_{n\rightarrow +\infty} f_n(x)=0$. This shows that the pointwise limit of $f_n$ is indeed the constant function equal to $0$ on $\mathbb{R}$.

Since $f_n(2n)=n$ for all $n$, the convergence is not uniform on $\mathbb{R}$.

Now let us fix an interval $[a,b]$. For all $n$ large enough ($n\geq \lfloor b\rfloor +1$), we have $f_n(x)=0$ for all $x\in [a,b]$. Hence $\sup_{x\in[a,b]}|f_n(x)|=0$ for these $n$ and the uniform convergence on $[a,b]$ follows.

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Let $x\in \mathbb R$. Then almost all $n\in \mathbb N$ are $>x$. Therefore, $f_n(x)=0$ for almost all $n$. Thus $f_n\to 0$ pointwise.

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