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) <br /> f_{n}(x) = \begin{cases} \frac{1}{x} &amp; , \frac{1}{n+1}\leq x &lt; 1 \\ 0 &amp; , \text{ otherwise } \end{cases}<br />

Homework Equations

\{ f_{n} \}_{n=1}^{\infty} is Cauchy in L^{1}[0,1] iff for all \epsilon&gt;0 there exists N\in\mathbb{N} such that for n,m\geq N, ||f_{n}-f_{m}||_{1} &lt; \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}} &amp; , \frac{1}{n+1}\leq x \leq 1 \\ 0 &amp; , \text{ otherwise } \end{cases}

(b) <br /> f_{n}(x) = \begin{cases} \frac{1}{x} &amp; , \frac{1}{n+1}\leq x &lt; 1 \\ 0 &amp; , \text{ otherwise } \end{cases}<br />

Homework Equations

\{ f_{n} \}_{n=1}^{\infty} is Cauchy in L^{1}[0,1] iff for all \epsilon&gt;0 there exists N\in\mathbb{N} such that for n,m\geq N, ||f_{n}-f_{m}||_{1} &lt; \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:
<br /> \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)),<br /> etc.

You can without loss of generality assume N \leq n &lt; 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&gt;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 &lt; \frac{1}{n+1}. So \displaystyle f_{m}-f_{n} = \begin{cases} \frac{1}{\sqrt{x}} &amp;, \frac{1}{m+1}\leq x &lt; \frac{1}{n+1} \\ 0 &amp;, \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:
<br /> \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)),<br /> etc.

You can without loss of generality assume N \leq n &lt; m.

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