Proving Simple Summable Functions and Cauchy Sequences on [0,1]

[Fermat1]

Let [x] be the integer part of x. Define the function $$f_{n}$$$$$$ by $$f_{n}=\frac{[nx]}{n}$$

1) show that every $$f_{n}$$ is a simple summable function.

So Firstly I need to show I can write is as a linear combination of indicator functions. Not sure how to proceed.

2)Show $$$$$$(f_{n}) $$is a cauchy sequence with the metric which is the integral from 0 to 1 of |f-g|.
I think the key to this question is what is the relation between the difference of the integer parts and the integer part of the difference.

3) show there is no simple summable f on [0,1] such that $$f_{n}$$ converges to f in the above metric

Thanks

 

[Fermat1]

Maybe define $$A_{n}=f^{-1}(n)$$ and $$t_{n}=n^{-1}$$

 

[Opalg]

Let [x] be the integer part of x. Define the function $$f_{n}$$ by $$f_{n}=\frac{[nx]}{n}$$

1) show that every $$f_{n}$$ is a simple summable function.

So Firstly I need to show I can write is as a linear combination of indicator functions. Not sure how to proceed.

2)Show $$(f_{n}) $$is a cauchy sequence with the metric which is the integral from 0 to 1 of |f-g|.
I think the key to this question is what is the relation between the difference of the integer parts and the integer part of the difference.

3) show there is no simple summable f on [0,1] such that $$f_{n}$$ converges to f in the above metric

Thanks

I think you have omitted the very important condition that these functions are supposed to be defined on the interval [0,1]. In that case, for 1) you should show that $f_n(x) = k/n$ when $k/n \leqslant x <(k+1)/n$, for $0\leqslant k\leqslant n-1.$

For 2), show that $x - \frac1n < f_n(x) \leqslant x$ (for $0\leqslant x\leqslant 1$) and deduce that $|f_n(x) - f_m(x)| <\frac1m$ whenever $n>m$. This shows that $f_n$ is uniformly Cauchy, and therefore Cauchy for the given metric.

For 3), $f_n(x) \to x$ uniformly on $[0,1]$. But the uniform metric is stronger than the given metric, so it follows that $f_n(x) \to x$ in the given metrix. But $x$ is not a simple function.

Intuitively, $f_n$ is a staircase function, going up from 0 to 1 in steps of height and width $1/n$.

 

[Fermat1]

Thanks. I will look at this evening and get back to you tomorrow.

 

[Fermat1]

I think you have omitted the very important condition that these functions are supposed to be defined on the interval [0,1]. In that case, for 1) you should show that $f_n(x) = k/n$ when $k/n \leqslant x <(k+1)/n$, for $0\leqslant k\leqslant n-1.$

For 2), show that $x - \frac1n < f_n(x) \leqslant x$ (for $0\leqslant x\leqslant 1$) and deduce that $|f_n(x) - f_m(x)| <\frac1m$ whenever $n>m$. This shows that $f_n$ is uniformly Cauchy, and therefore Cauchy for the given metric.

For 3), $f_n(x) \to x$ uniformly on $[0,1]$. But the uniform metric is stronger than the given metric, so it follows that $f_n(x) \to x$ in the given metrix. But $x$ is not a simple function.

Intuitively, $f_n$ is a staircase function, going up from 0 to 1 in steps of height and width $1/n$.

Is there a formula for [x]?

 

[Opalg]

Is there a formula for [x]?

The only formula is that if there is an integer $m$ such that $m\leqslant x<m+1$ then $\lfloor x\rfloor = m.$

 

[Fermat1]

I think you have omitted the very important condition that these functions are supposed to be defined on the interval [0,1]. In that case, for 1) you should show that $f_n(x) = k/n$ when $k/n \leqslant x <(k+1)/n$, for $0\leqslant k\leqslant n-1.$

For 2), show that $x - \frac1n < f_n(x) \leqslant x$ (for $0\leqslant x\leqslant 1$) and deduce that $|f_n(x) - f_m(x)| <\frac1m$ whenever $n>m$. This shows that $f_n$ is uniformly Cauchy, and therefore Cauchy for the given metric.

For 3), $f_n(x) \to x$ uniformly on $[0,1]$. But the uniform metric is stronger than the given metric, so it follows that $f_n(x) \to x$ in the given metrix. But $x$ is not a simple function.

Intuitively, $f_n$ is a staircase function, going up from 0 to 1 in steps of height and width $1/n$.

---

 

[Fermat1]

Ok the maths code has gone haywire for some reason. Basically, in trying to write $$f_{n}$$ as a linear combination of indicator functions, I came up with a choice which meant $$f_{n}$$$$$$ was not summable. By the way by summable I mean that the series of $|t_{k}|u(A_{k})$ converges, where u is the measure and the $t_{k}$ ,$A_{k}$ are the real numbers and measurable sets in the linear combination .

The other 2 questions I have completed.

 

[Fermat1]

I put $$t_{m}=\frac{m}{n}$$ and $A_{m}$=[m/n,(m+1)/n)

 

[Opalg]

I put $$t_{m}=\frac{m}{n}$$ and $A_{m}$=[m/n,(m+1)/n)

(Yes)

 

[Fermat1]

(Yes)

But $$u(A_{m})=\frac{1}{n}$$ so $$t_{m}u(A_{m})=\frac{m}{n^2}$$, the series of which from m=1 to infinity does not converge. Is it that since the domain is [0,1], m only goes up to n, so it is in fact a finite sum?

 

[Opalg]

But $$u(A_{m})=\frac{1}{n}$$ so $$t_{m}u(A_{m})=\frac{m}{n^2}$$, the series of which from m=1 to infinity does not converge. Is it that since the domain is [0,1], m only goes up to n, so it is in fact a finite sum?

That is what made me think that the domain must be the unit interval. The functions $f_n$ are certainly not summable over the whole real line. Also, question 2) refers to "the metric which is the integral from 0 to 1 of |f-g|", another indication that the domain should be the unit interval.