1
$\begingroup$

Take the measure space to be $\mathbb{R}$ with Borel $\sigma$-algebra and Lebesgue measure (Although just thinking in terms of a general measure space probably works for this problem.)

Question: True or False

If $1\leq p<\infty$, $f_n\geq1$, and $f_n\rightarrow f$ in $L_p$, then $\frac{1}{f_n}\rightarrow\frac{1}{f}$ in $L_p$.

I couldn't come up with a counter example, but I'm also not able to prove the statement.

Things that may help are convergence in $L_p$ implies convergence in measure and convergence of subsequences almost uniform and almost everywhere. I tried combining fractions and using the fact that $f_n\geq1$ to string together inequalities but I wasn't able to get the desired convergence.

It is quite possible that I am missing something simple. Any Guidance?

$\endgroup$
3
  • $\begingroup$ I guess I was confused on that. When we talk about convergence in $L_p$ we are implying that $f_n$ and $f$ are in L_p. Otherwise $\int |1-1|^pd\lambda$ is always zero, but that doesn't mean the constant function 1 converges to 1 in $L_p$ on an infinite measure space correct? $\endgroup$ Commented Jul 28, 2015 at 0:57
  • $\begingroup$ So the statement stats off as false when they claim $f_n\rightarrow f$ in $L_p$?. Thanks for the help I was confused by that in the beginning. $\endgroup$ Commented Jul 28, 2015 at 1:01
  • $\begingroup$ True, they had a the disclaimer at the top of the exam to assume Lebesgue measure on real numbers unless otherwise specified. The finite measure case is a lot more interesting though and perhaps that is what was intended. At least the problem made me read over the definition of convergence in Lp more closely. Again thanks. $\endgroup$ Commented Jul 28, 2015 at 1:09

1 Answer 1

1
$\begingroup$

From $f_n - f\to 0$ in $L^p$ follows that there exists a subsequence of $f_n$ which converges almost everywhere to $f$. Thus, $f\ge 1$ almost everywhere.

Now, notice that

$$ \left| \frac 1 {f_n} - \frac1f \right | = \frac{|f_n - f|}{f_n f} \le |f_n - f|. $$

Integrate (take essential supremum) on both side to obtain the statement.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .