- #1

Euler2718

- 90

- 3

## Homework Statement

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

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

(b) [tex]

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

[/tex]

## Homework Equations

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

## The Attempt at a Solution

[/B]

In both problems I

*think*I should get to a step where I integrate [itex]\left|\frac{1}{\sqrt{x}}\right| [/itex] and [itex]\left|\frac{1}{x}\right| [/itex] over [itex][0,1][/itex] and get the norm values of [itex]2[/itex] and [itex]undefined[/itex] respectively, then I can conclude easily. Embarrassingly I don't know what [itex]f_{n}-f_{m}[/itex] 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 [itex]f_{n}[/itex] and [itex]f_{m}[/itex] in either case ?