Sequence function on $C[-2,2]$ is Cauchy Sequence

Question

We consider the space $C[-2,2]$ and Euclidian norm $\left\| . \right\|_2$. We observe the sequence of functions:

$$ f_n(x)=\left\{\begin{matrix}0, & x \notin [0,1]\\ nx, & x \in [0,\frac{1}{n}]\\ 1, & x \in [\frac{1}{n},1- \frac{1}{n}] \\ n(1-x) & x \in [1-\frac{1}{n},1] \end{matrix}\right. $$

Is $f_n(x)$ a Cauchy Sequence wrt $\left\| . \right\|_2$? I am struggile to show either of those things. I cannot find counter example and I don´t get far when I try to prove that it is. Any tip or help would be appriciate!

Answer 1

As $n$ increases, $f_n$ resembles a so-called rectangular function (see also plots below). So you can have the intuition that somehow it "converges" and could be a Cauchy sequence.

To prove it, you would have to estimate $\|f_n - f_m\|_{L^2}$ for large $n,m$. Since the computations get tedious, an idea suggested above is to use the triangular inequality and write $$ \| f_n - f_m \|_{L^2} \leq \| f_n - F \|_{L^2} + \| F - f_m \|_{L^2} $$ where $F$ is the "limit" rectangular function $F(x) = 1$ on $(0,1)$ and $F(x) = 0$ on $[-2,0]\cup[1,2]$.

Now to prove that $\|f_n-F\|_{L^2} \to 0$ as $n \to +\infty$, you can either use the dominated convergence theorem as in the other answer, or perform the computation by hand. Indeed, $$ \| f_n - F \|_{L^2(-2,2)}^2 = \int_0^1 |f_n(x) - 1|^2 \mathrm{d} x = 2 \int_0^{\frac 1 n} |n x - 1|^2 \mathrm{d}x = \frac 2 n \int_0^1 |y -1|^2 \mathrm{d} y \to 0. $$

From this, you conclude that, for every $\varepsilon > 0$, there exists $N_0 \in \mathbb{N}$ such that, for $n,m \geq N_0$, $\|f_n-f_m\|_{L^2} \leq \varepsilon$, so you indeed have a Cauchy sequence.

The "interesting" part of your exercise is that the limit function $F$ is not continuous on $[-2,2]$. This is not a contradiction, because you have proved that the sequence is a Cauchy sequence for the $\|\cdot\|_{L^2}$ norm, not for the $\|\cdot\|_{C^0}$ one.

Here is the Python script that was used to produce this plot.

import numpy as np
import matplotlib.pyplot as plt

def generate_fn_values(N):
    x = np.linspace(-2, 2, 1000)
    y = np.piecewise(x, [(x >= -2) & (x <= 0), (x > 0) & (x <= 1/N), (x > 1/N) & (x <= 1-1/N), (x > 1-1/N) & (x <= 1), (x > 1) & (x <= 2)], [0, lambda x: N * x, 1, lambda x: N * (1-x), 0])
    return x, y

x3, y3 = generate_fn_values(3)
x4, y4 = generate_fn_values(4)
xN, yN = generate_fn_values(20)

plt.plot(x3, y3, label="f_3")
plt.plot(x4, y4, label="f_4")
plt.plot(xN, yN, label="f_20")

plt.xlabel("x-axis")
plt.ylabel("y-axis")
plt.title("Functions f_3 and f_4 and f_20")
plt.legend()
plt.show()

Answer 2

If you are considering the $L^{2}-$ norm then the sequence is Cauchy. Note that $0\leq f_n \leq 1$ amd $f_n (x) \to 1$ for $0<x<1$. By DCT we get $\int |f_n-1|^{2} \to 0$. So $(f_n)$ is actually convergent.