Showing a sequence of functions is Cauchy/not Cauchy in L1

[Euler2718]

Homework Statement

Determine whether or not the following sequences of real valued functions are Cauchy in $L^{1}[0,1]$:

(a) $$f_{n}(x) = \begin{cases} \frac{1}{\sqrt{x}} & , \frac{1}{n+1}\leq x \leq 1 \\ 0 & , \text{ otherwise } \end{cases}$$

(b) $$f_{n}(x) = \begin{cases} \frac{1}{x} & , \frac{1}{n+1}\leq x < 1 \\ 0 & , \text{ otherwise } \end{cases}$$

Homework Equations

$\{ f_{n} \}_{n=1}^{\infty}$ is Cauchy in $L^{1}[0,1]$ iff for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that for $n,m\geq N$, $||f_{n}-f_{m}||_{1} < \epsilon$.

The Attempt at a Solution

[/B]
In both problems I think I should get to a step where I integrate $\left|\frac{1}{\sqrt{x}}\right|$ and $\left|\frac{1}{x}\right|$ over $[0,1]$ and get the norm values of $2$ and $undefined$ respectively, then I can conclude easily. Embarrassingly I don't know what $f_{n}-f_{m}$ is explicitly (in terms of its piecewise definition). I had first thought it to be zero for both cases, I don't think this is the case. Feel stupid asking this, but how do you subtract $f_{n}$ and $f_{m}$ in either case ?

 

[Ray Vickson]

Homework Statement

Determine whether or not the following sequences of real valued functions are Cauchy in $L^{1}[0,1]$:

(a) $$f_{n}(x) = \begin{cases} \frac{1}{\sqrt{x}} & , \frac{1}{n+1}\leq x \leq 1 \\ 0 & , \text{ otherwise } \end{cases}$$

(b) $$f_{n}(x) = \begin{cases} \frac{1}{x} & , \frac{1}{n+1}\leq x < 1 \\ 0 & , \text{ otherwise } \end{cases}$$

Homework Equations

$\{ f_{n} \}_{n=1}^{\infty}$ is Cauchy in $L^{1}[0,1]$ iff for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that for $n,m\geq N$, $||f_{n}-f_{m}||_{1} < \epsilon$.

The Attempt at a Solution

[/B]
In both problems I think I should get to a step where I integrate $\left|\frac{1}{\sqrt{x}}\right|$ and $\left|\frac{1}{x}\right|$ over $[0,1]$ and get the norm values of $2$ and $undefined$ respectively, then I can conclude easily. Embarrassingly I don't know what $f_{n}-f_{m}$ is explicitly (in terms of its piecewise definition). I had first thought it to be zero for both cases, I don't think this is the case. Feel stupid asking this, but how do you subtract $f_{n}$ and $f_{m}$ in either case ?

If $m > n$, in what $x$-region are $f_m(x)$ and $f_n(x)$ the same? In what $x$-region are they different? So, what is the formula for $f_m(x) - f_n(x)$ over $0 \leq x \leq 1?$

 

[pasmith]

Your functions are defined piecewise, and vanish on $[0, \frac{1}{n+1})$. You shouldn't be getting any undefined integrals:
$$\int_0^1 f_n(x)\,dx = \int_0^{1/(n+1)} 0\,dx + \int_{1/(n+1)}^1 \frac1x\,dx = \left[\log(x)\right]_{1/(n+1)}^1 = -\log(1/(n+1)),$$ etc.

You can without loss of generality assume $N \leq n < m$.

 

[Euler2718]

If $m > n$, in what $x$-region are $f_m(x)$ and $f_n(x)$ the same? In what $x$-region are they different? So, what is the formula for $f_m(x) - f_n(x)$ over $0 \leq x \leq 1?$

Using (a) as an example, if $m>n$, then $f_{m}$ and $f_{n}$ would be the same (both $\frac{1}{\sqrt{x}}$ ) on $\frac{1}{n+1} \leq x \leq 1$ and different ($f_{m}=\frac{1}{\sqrt{x}}$ but $f_{n}=0$) on $\frac{1}{m+1}\leq x < \frac{1}{n+1}$. So $\displaystyle f_{m}-f_{n} = \begin{cases} \frac{1}{\sqrt{x}} &, \frac{1}{m+1}\leq x < \frac{1}{n+1} \\ 0 &, \text{ otherwise } \end{cases}$ ?

Your functions are defined piecewise, and vanish on $[0, \frac{1}{n+1})$. You shouldn't be getting any undefined integrals:
$$\int_0^1 f_n(x)\,dx = \int_0^{1/(n+1)} 0\,dx + \int_{1/(n+1)}^1 \frac1x\,dx = \left[\log(x)\right]_{1/(n+1)}^1 = -\log(1/(n+1)),$$ etc.

You can without loss of generality assume $N \leq n < m$.

I see what you mean. I was not careful enough with the permissible $x$ values to notice this.