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I have a basic problem understanding the convergence of the "tent" function

\begin{equation} x_n(r) = \begin{cases} nr,\; r \in \left[0, \frac{1}{n}\right]\\ 2 - nr, \; r \in \left[\frac{1}{n}, \frac{2}{n}\right]\\ 0,\; r\in \left[\frac{2}{n}, 1\right] \end{cases} \end{equation}

to the zero function, $x_0(r)=0, \; \forall r\in [0, 1]$ pointwise in $r$.

I understand that the function concentrates around zero in the ordinates that depend on $n$, but doesn't it still place a lot of mass at that point?

Also, why is the convergence not uniform using the $\sup$-metric?

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If $r=0$ then for all $n$, $x_n(r)=0$. If $r\neq 0$ we can find an integer $n_0$ such that $\frac 2n\leq x_0$ for $n\geq n_0$, hence $x_n(r)=0$ for $n\geq n_0$. Since for all $n$ we have $\sup_{r\in\left[0,1\right]}|x_n(r)-0|=1$, the convergence cannot be uniform. – Davide Giraudo Sep 7 '11 at 9:55
By "that point", do you mean zero? The function doesn't place any "mass" there; the value at zero is zero. – joriki Sep 7 '11 at 10:10
@davide thanks for your patient answer.@joriki, quite so. :) – Praetoria Sep 7 '11 at 10:29
up vote 1 down vote accepted

In order to expand a little my comment, I draw some functions $x_n$.

Some $x_n$

We notice that $x_n(0)=0$ for all $n$ and if $r\in\left(0,1\right]$, then for $n\geq \lfloor x\rfloor+1$, where $\lfloor\cdot\rfloor$ is the floor function, we get $x_n(r)=0$, hence $\displaystyle \lim_{n\to +\infty}x_n(r)=0$ for all $r\in\left[0,1\right]$ and the sequence $\{x_n\}$ converges pointwise to the zero function. But the convergence is not uniform. An intuitive way to see that is that, for example $\varepsilon=\frac 12$, we get for each $r$ a smallest $n_0(r)$ for which $|x_n(r)|\leq \frac 12$ for $n\geq n_0(r)$, but this $n_0(r)$ cannot be chosen independently of $r$. A more formal argument is the following: since $x_n\left(\frac 1n\right)=1$ we have $\sup_{r\in\left[0,1\right]}|x_n(r)-0|=1$ and the convergence cannot be uniform on $\left[0,1\right]$ (but of course it is on each interval $\left[a,1\right]$ with $0<a<1$, since for $n$ large enough we have $x_n(r)=0$ for all $r\in\left[a,1\right]$).

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