Consider the metric space $(C[0,1],d_1).$ Consider the sequence of piecewise linear functions:
\begin{align}
f_n(x) = \left\{\begin{matrix}
0 & 0 \leq x \leq \frac{1}{2}-\frac{1}{n}\\
\frac{1}{2} – \frac{n}{4} + \frac{n}{2}x & \,\,\,\,\,\, \frac{1}{2} – \frac{1}{n} <x<\frac{1}{2} + \frac{1}{n},
\,\,\,\, n \geq 2.\\
1 & \frac{1}{2} + \frac{1}{n} \leq x \leq 1.
\end{matrix}\right.
\end{align}
I want to determine if the following sequence is Cauchy and if it converges in $d_1$ metric to any function in $C[0,1].$ Recall the $d_1$ metric in $C[0,1]$ :
$$d_1(f,g) = \int_{0}^{1}|f(x) – g(x)|dx$$
$\mathbf{My\,attempt}$:
After sketching first few functions in the sequence, I think this is a Cauchy sequence. In order to prove this rigorously, for all $\varepsilon>0$, I must find $K(\varepsilon)$ such that $d_1(f_n,f_m)<\varepsilon$ whenever $m,n ≥ K(\varepsilon).$ Here is my attempt:
Assuming $n > m,$
\begin{align}
d_1(f_n,f_m) &= \int_{0}^{1}|f_n(x) – f_m(x)|dx \\
&= \frac{(n-m)}{4}\int|2x-1|dx \\
&\leq \frac{(n-m)}{4}.width.\sup_{x \in I}|2x-1|\\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots &\\
&\leq \varepsilon.
\end{align}
I can't seem to work out integration bounds in the second step and that's why I'm unable to compute the width of the interval. Any help highly appreciated on how to fix this.
For the convergence part, my guess is that the sequence converges point-wise to to a discontinous step function. Thus, it doesn't converge in $d_1$ metric to any function in $C[0,1].$ But how can I prove this rigorously? Please note that I want to show this directly. I don't want to refer to any larger spaces such as $L^p$ spaces.
Best Answer
Assume $m > n$. From your sketching it should be clear (and it is easy to prove that) $f_m(x) = f_n(x)$ except on the open interval $I_n = (\frac{1}{2} - \frac{1}{n}, \frac{1}{2} + \frac{1}{n})$. $|f_m(x) - f_n(x)| \le 1$ on $I_n$. So: $$ \begin{align*} d_1(f_m, f_n) &\le \int_0^{\frac{1}{2} - \frac{1}{n}} 0dx + \int_{\frac{1}{2} - \frac{1}{n}}^{\frac{1}{2} + \frac{1}{n}} 1 dx + \int_{\frac{1}{2} + \frac{1}{n}}^1 0dx \\ &= \frac{2}{n} \end{align*} $$ and you can take $K(\epsilon)$ to be any $n$ such that $\frac{2}{n} < \epsilon$
The sequence converges pointwise to the function $f$ defined by: $$ f(x) = \left\{ \begin{array}{l@{\quad}l} 0 &\mbox{if $0 \le x < \frac{1}{2}$}\\ \frac{1}{2}&\mbox{if $x = \frac{1}{2}$}\\ 1&\mbox{if $\frac{1}{2} < x \le 1$}\\ \end{array}\right. $$ You can see this by observing that, (A), for $x \in [0, 1] \setminus\{\frac{1}{2}\}$, $x \not\in I_n$ for all sufficiently large $n$ (so that $f_n(x) = t$ for all sufficiently large $n$ , where $t = 0$ if $x < \frac{1}{2}$ and $t = 1$ if $x > \frac{1}{2}$), while, (B), $f_n(\frac{1}{2}) = \frac{1}{2}$ for all $n$.