# Does convergence in $L_p$ imply convergence of quotients in $L_p$

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?

• 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? Commented Jul 28, 2015 at 0:57
• 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. Commented Jul 28, 2015 at 1:01
• 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. Commented Jul 28, 2015 at 1:09

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