# Determine if the following sequence is Cauchy and converges in $(C[0,1],d_1)?$

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}

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.

• estimate the step $d_1(f_{n+1},f_n)$ Jun 23 '21 at 22:29
• If you try to draw your functions, you might get a better intuition for what it's converging to. It should be clear that this is converging to a step function $f(x)=\begin{cases}0&x\leq 1/2\\1& x> 1/2\end{cases}$ (I am saying a step-function because the value at $1/2$ is ambigious.) That being said $f$ is not in $C[0,1]$. Jun 23 '21 at 22:49
• Regarding the Cauchy part, I think you may need to be a bit careful with the tails at $1/2\pm 1/n$ and $1/2\pm 1/m$. Actually, you're integral gets quite messy if you try to split it up. It's probably easier to compute the triangles bounded by $f_m$ and $f_n$. You should get two triangles of base $|1/n-1/m|$ and height $1/2$. Jun 23 '21 at 22:54
• @daruma why is the value at $1/2$ ambiguous and could you please elaborate how you got two triangles of base $|1/n - 1/m|$ and height $1/2$?
– user715112
Jun 24 '21 at 2:01

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 $$n$$.
• Assuming $m >n,$ it is not clear to me from the sketches as to why $f_m(x) = f_n(x)$ except on $I_n$? Also, how did you establish that $f_n(\frac{1}{2}) = \frac{1}{2} \text { for all}\,\, n ?$
• For point (1) if $m > n$, then $f_m(x) - f_n(x)$ for $0 \le x \le 1/2 - 1/n$ and for $1/2 + 1/n \le x \le 1$, because $1/n > 1/m$. For point (2) , you have $f_n(1/2) = 1/2 - n/4 + (n/2)(1/2) = 1/2$ for all $n$. Jun 24 '21 at 2:25
• I see what you mean: $f_m(x) - f_n(x)$ is a typo for $f_m(x) = f_n(x)$ in my comment. Jun 24 '21 at 3:02