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[SOLVED] Dominated convergence theorem

Fermat

Active member
Nov 3, 2013
188
Define \(\displaystyle f_{n}(x)=\frac{n^{1.5}x}{1+n^{2}x^2}\) for x in [0,1]. Use Dominated convergence theorem to find the limit of the integral of f_n over [0,1].

I find that f_n converges to 0 so if I can find domination function I have shown integral is zero. Correct? I find f_n is dominated by function g where g(x)=x^-2 when x is not zero and g(0)=0. Is such a function integrable?

Thanks
 
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Fermat

Active member
Nov 3, 2013
188
Re: dominated convergence theorem

Just noticed the code isn't displaying, sorry about that. I can't see whats wrong though.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: dominated convergence theorem

Define \(\displaystyle f_{n}(x)=\frac{n^{1.5}x}{1+n^{2}x^2}\) for x in [0,1]. Use Dominated convergence theorem to find the limit of the integral of f_n over [0,1].

I find that f_n converges to 0 so if I can find domination function I have shown integral is zero. Correct? I find f_n is dominated by function g where g(x)=x^-2 when x is not zero and g(0)=0. Is such a function integrable?
That is the right approach, but the function $g(x) = x^{-2}$ will not do the job. It certainly dominates the functions $f_n$ but it is not integrable over the interval $[0,1].$ If you were using the Riemann integral, you would say that the improper integral \(\displaystyle \int_0^1 g(x)\,dx\) diverges because \(\displaystyle \int_{\varepsilon}^1x^{-2}dx = 1- \varepsilon^{-1}\), which goes to $\infty$ as $\varepsilon\searrow0.$ The same calculation shows that $g(x)$ is not Lebesgue integrable either.

So you need to find a "better" dominating function, and I think that you should look at $g(x) = x^{-1/2}$ (for $x\ne0$, with $g(0)=0$). To see that this function dominates $f_n$, notice that \(\displaystyle \frac{n^{1.5}x}{1+n^{2}x^2} \leqslant \frac1{\sqrt x}\:\Leftrightarrow\: n^{1.5}x^{1.5} \leqslant 1+n^2x^2.\) Put $y = \sqrt{nx}$ and use elementary calculus to show that $y^3 \leqslant 1+y^4$ for all positive $y.$

[The problem with the LaTeX code was that there was an extra } in it. The LaTeX compiler gets completely thrown by non-matching braces.]
 

Fermat

Active member
Nov 3, 2013
188
Re: dominated convergence theorem

That is the right approach, but the function $g(x) = x^{-2}$ will not do the job. It certainly dominates the functions $f_n$ but it is not integrable over the interval $[0,1].$ If you were using the Riemann integral, you would say that the improper integral \(\displaystyle \int_0^1 g(x)\,dx\) diverges because \(\displaystyle \int_{\varepsilon}^1x^{-2}dx = 1- \varepsilon^{-1}\), which goes to $\infty$ as $\varepsilon\searrow0.$ The same calculation shows that $g(x)$ is not Lebesgue integrable either.

So you need to find a "better" dominating function, and I think that you should look at $g(x) = x^{-1/2}$ (for $x\ne0$, with $g(0)=0$). To see that this function dominates $f_n$, notice that \(\displaystyle \frac{n^{1.5}x}{1+n^{2}x^2} \leqslant \frac1{\sqrt x}\:\Leftrightarrow\: n^{1.5}x^{1.5} \leqslant 1+n^2x^2.\) Put $y = \sqrt{nx}$ and use elementary calculus to show that $y^3 \leqslant 1+y^4$ for all positive $y.$

[The problem with the LaTeX code was that there was an extra } in it. The LaTeX compiler gets completely thrown by non-matching braces.]
Thanks Opalg. On a seperate note can you tell me what the epsilon/2^n or 1/2^n 'trick' is please?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: dominated convergence theorem

Thanks Opalg. On a separate note can you tell me what the epsilon/2^n or 1/2^n 'trick' is please?
No, I don't know of any trick with those names. Maybe someone else here does?