Let $(X,\mu)$ be a measure space with $\mu(X)=1$. Let $f$ be a measurable function such that $0<||f||_{\infty}< \infty$. Let $a_n=\int|f|^nd\mu$

Show that:

(1) $\lim_{n\to\infty}a_{n+1}/a_n=||f||_\infty$

(2) $\lim_{n\to\infty}a_n^{1/n}=||f||_\infty$

I have no idea how to proceed.

  • 2
    $\begingroup$ For the second one : you can see here. $\endgroup$ – Watson Feb 15 '16 at 15:34
  • $\begingroup$ For the first one, I am not even sure the question is defined.What if $a_n=0$, say for instance for $f=\mathbb{1}_{\mathbb{Q}}$? $\endgroup$ – Clement C. Feb 15 '16 at 15:48
  • $\begingroup$ @ClementC. $\| \mathbf{1}_\mathbb{Q} \|_\infty = 0$, assuming that $\| \cdot \|_\infty$ is the essential supremum. $\endgroup$ – Nigel Overmars Feb 15 '16 at 15:53
  • $\begingroup$ @NigelOvermars granted, but this "assuming that" is not really part of the question. $\endgroup$ – Clement C. Feb 15 '16 at 15:54
  • $\begingroup$ @Clement: In a measure theory context, $\|f\|_\infty $ is nearly always the essential supremum of $|f|$. $\endgroup$ – PhoemueX Feb 15 '16 at 15:59

By renormalizing, we can assume $\left\Vert f\right\Vert _{\infty}=1$. Set $a_{n}:=\int\left|f\right|^{n}\,{\rm d}\mu$ and $b_{n}:=\frac{a_{n+1}}{a_{n}}$ for $n\in\mathbb{N}$. Then $\left\Vert f\right\Vert _{\infty}$ implies $a_{n+1}\leq a_{n}$ and thus $b_{n}\leq1$ for all $n\in\mathbb{N}$.

Now, the Cauchy-Schwarz inequality implies that $$ a_{n+1}=\int\left|f\right|^{n+1}\,{\rm d}\mu=\int\left|f\right|^{\frac{n}{2}}\cdot\left|f\right|^{\frac{n+2}{2}}\,{\rm d}\mu\leq\sqrt{\int\left|f\right|^{n}\,{\rm d}\mu}\cdot\sqrt{\int\left|f\right|^{n+2}\,{\rm d}\mu}, $$ i.e. $a_{n+1}^{2}\leq a_{n}\cdot a_{n+2}$. For the quotient $b_{n}=\frac{a_{n+1}}{a_{n}}$, this implies $$ b_{n}=\frac{a_{n+1}}{a_{n}}\leq\frac{a_{n+2}}{a_{n+1}}=b_{n+1}, $$ so that $\left(b_{n}\right)_{n\in\mathbb{N}}$ is nondecreasing and bounded above by $1$. Thus, $b:=\lim_{n\to\infty}b_{n}=\sup_{n\in\mathbb{N}}b_{n}\in\left(0,1\right]$ exists.

Now, the ratio test easily implies that the power series $$ \sum_{n=1}^{\infty}a_{n}x^{n} $$ has radius of convergence $R=\frac{1}{b}$.

But as noted in the comments, the question Limit of $L^p$ norm shows that we have $$ \sqrt[n]{a_{n}}\xrightarrow[n\to\infty]{}\left\Vert f\right\Vert _{\infty}=1. $$ Using the root test, we see that the power series from above also has the radius of convergence $R=\frac{1}{\lim_{n\to\infty}\sqrt[n]{a_{n}}}=1$. All in all, we conclude $\lim_{n\to\infty}b_{n}=R=1=\left\Vert f\right\Vert _{\infty}$, as desired.

EDIT: What I am using here is in more detail as follows: From above, we know that the limit $b := \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ exists.

Now, if $|x| < 1/b$, we have $$ \bigg|\frac{a_{n+1}x^{n+1}}{a_n x^n}\bigg| = \frac{a_{n+1}}{a_n} \cdot |x| \to b |x| <1, $$ so that the series $\sum_n a_n x^n$ converges (absolutely).

Conversely, if $|x|> 1/b$, a similar argument shows that the series diverges. Hence, the radius of convergence is given by $1/b$.

Essentially the same holds for the root test.

  • $\begingroup$ The part of proof that the limit $a_{n+1}/a_{n}$ is incorrect using root and ration tests on power series. at the boundary of both tests, the result should be inconclusive. $\endgroup$ – runaround Feb 19 '16 at 19:14
  • $\begingroup$ @runaround: Yes, at the boundary, it is inconclusive. Nevertheless, the radius of convergence is given by $\lim_n |a_{n}/a_{n+1}|$ if that limit exists. Likewise for the root test. See e.g. here: en.wikipedia.org/wiki/… $\endgroup$ – PhoemueX Feb 19 '16 at 19:27
  • $\begingroup$ @PhiemueX, if the power serie has the convergence radius 1, it does not give the value of the convergence radius from the ratio test is also 1. $\endgroup$ – runaround Feb 19 '16 at 19:50
  • $\begingroup$ @runaround: Please see my edit. $\endgroup$ – PhoemueX Feb 19 '16 at 20:48

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