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Let $f_n(x)=x^{n+1/2}$ for $n\geq 1$ and $x\in [-1,1]$. Show that for $|a|<1$ the series $\sum_{n=1}^{\infty}a^nf_n$ converges in $L_2([-1,1])$. First, I have shown that $\sum_{n=1}^{\infty}a^nf_n(x)$ converges to $\frac{ax^{3/2}}{1-ax}$ for $x\in [-1,1]$. After that, I'm having difficulties to show that $$ \int_{-1}^{1}\left | \frac{ax^{3/2}}{1-ax} \right |^2\,\mathrm{d}x=|a|^2\int_{-1}^{1}\frac{|x^{3}|}{\left | 1-ax \right |^2}\,\mathrm{d}x $$ is less than $+\infty$. The denominator in the integral is troublesome.

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  • $\begingroup$ Isn't there a problem with $x^{n+1/2}$ for $x<0?$ $\endgroup$ – zhw. Jan 10 at 23:35
  • $\begingroup$ @zhw. You are right, I've only considered it on $[0, 1]$. I'll talk to my lecturer. $\endgroup$ – UnknownW Jan 10 at 23:40
  • $\begingroup$ @zhw. It is a function from $[-1,1]$ into $\mathbb{C}$, so it's all right. For $x<0$, we may put $\sqrt{x}=i\sqrt{-x}$. $\endgroup$ – UnknownW Jan 11 at 15:33
  • $\begingroup$ That's unusual, so you should probably mention it in your question. $\endgroup$ – zhw. Jan 11 at 19:42
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Since $|a|<1$ and in the integral $|x|<1$, we have $(1-ax)>0$, so

$$|a|^2\int_{-1}^1 \frac{|x^3|}{|1-ax|^2}dx = a^2\int_0^1 \frac{x^3}{(1-ax)^2}dx\,-a^2\int_{-1}^0\frac{x^3}{(1-ax)^2}dx.$$

Substituting $u=1-ax$ should finish the job.

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Hint: The denominator is not troublesome; it can never be $0$ on the given interval. Another thing: If $\sum \|g_n\|_2 <\infty,$ then $\sum g_n$ converges in $L^2.$

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  • $\begingroup$ What does $||\cdot ||_2$ mean? $\endgroup$ – UnknownW Jan 15 at 22:31
  • $\begingroup$ It's the $L^2$ norm $\endgroup$ – zhw. Jan 15 at 22:58

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