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Simple summable functions

Fermat

Active member
Nov 3, 2013
188
Let [x] be the integer part of x. Define the function \(\displaystyle f_{n}\) by \(\displaystyle f_{n}=\frac{[nx]}{n}\)

1) show that every \(\displaystyle f_{n}\) 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 \(\displaystyle (f_{n}) \)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 \(\displaystyle f_{n}\) converges to f in the above metric

Thanks
 

Fermat

Active member
Nov 3, 2013
188
Maybe define \(\displaystyle A_{n}=f^{-1}(n)\) and \(\displaystyle t_{n}=n^{-1}\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Let [x] be the integer part of x. Define the function \(\displaystyle f_{n}\) by \(\displaystyle f_{n}=\frac{[nx]}{n}\)

1) show that every \(\displaystyle f_{n}\) 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 \(\displaystyle (f_{n}) \)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 \(\displaystyle f_{n}\) 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 $1/n$.
 

Fermat

Active member
Nov 3, 2013
188
Thanks. I will look at this evening and get back to you tomorrow.
 

Fermat

Active member
Nov 3, 2013
188
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
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 $\lfloor x\rfloor = m.$
 

Fermat

Active member
Nov 3, 2013
188
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 $1/n$.
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Last edited:

Fermat

Active member
Nov 3, 2013
188
Ok the maths code has gone haywire for some reason. Basically, in trying to write \(\displaystyle f_{n}\) as a linear combination of indicator functions, I came up with a choice which meant \(\displaystyle f_{n}\) 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.
 

Fermat

Active member
Nov 3, 2013
188
I put \(\displaystyle t_{m}=\frac{m}{n}\) and $A_{m}$=[m/n,(m+1)/n)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705

Fermat

Active member
Nov 3, 2013
188
But \(\displaystyle u(A_{m})=\frac{1}{n}\) so \(\displaystyle t_{m}u(A_{m})=\frac{m}{n^2}\), 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

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
But \(\displaystyle u(A_{m})=\frac{1}{n}\) so \(\displaystyle t_{m}u(A_{m})=\frac{m}{n^2}\), 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.