# If $f_n→f$ in $L^p(E)$, does that imply that $(f_n)^p→f^p$ in $L^1(E)$?

The question is just like the title. (For $E$ measurable and $1\le p<∞$, define $L^p(E)$ to be the collection of measurable functions $f$ for which $|f|^p$ is integrable over $E$; thus $L^1(E)$ is the collection of integrable functions.)

-
-1: This question does not show any research effort. –  Jesse Madnick Dec 2 '12 at 11:03
This is your 18th question on this site. To my understanding, not one of your 18 questions contains anything more than the problem statement. At the very least you could say where you found the problem or why you care about its solution. –  Jesse Madnick Dec 2 '12 at 11:06

Let $X = [1, \infty]$, $f_n(x) = \dfrac{1}{nx} + 1$ and $f(x) = 1$. We have:

$$\|f_n - f\|_2 = \left\{\int_1^\infty \left|\dfrac{1}{nx}\right|^2\,dx\right\}^{1/2} = \dfrac{1}{n}$$

Therefore, $f_n \to f$ in $L^2([1, \infty])$.

On the other hand:

\begin{align*} \|f_n^2 - f^2\|_1 &= \int_1^\infty \left|\left(\dfrac{1}{nx} + 1\right)^2 - 1\right| \,dx \\ &= \int_1^\infty \left|\left(\dfrac{1}{nx}\right)^2 + \dfrac{2}{nx}\right| \,dx \\ &\ge \dfrac{2}{n} \int_1^\infty \dfrac{1}{x} \,dx \end{align*}

Which diverges no matter what $n$ is.

-
In your answer n->∞, 1/n->0, so it's not divergent. I am sorry that the problem is true (don't try to find a counterexample - sorry for those who wasted their lives in a fruitless search) and there are some hints: First prove that for x,y >= 0, |x^p - y^p| <= p (x + y)^(p-1) |x-y| (you will need to use, in part, the mean value theorem). Holder's inequality will also prove useful... But who can prove this question? –  i_a_n Dec 4 '12 at 6:26
@i_a_n The points in your comment are simply not true. You have to choose a finite $n \in \mathbb{N}$ if you want to show convergence. And no matter what $n$ you choose, the RHS will always be $\infty$. As for your approach, it assumes $(x + y)^{p-1}$ is bounded if I understood correctly, but this doesn't have to be the case. –  Ayman Hourieh Dec 4 '12 at 11:00
In a question of this type, always check $f(x)=1/x$ in the sets $E_1=(0,1)$ and $E_2=(1,\infty)$. Notice that $f(x)$ is not integrable on either set in $L^1$, but it is integrable on one of them in $L^p$.
I think the question would be whether $1/x^p$ is in $L^1(E_2)$ when $1/x$ is in $L^p(E_2)$. –  jathd Dec 2 '12 at 9:26
If $E \subset \mathbb R^n$ is bounded, this is true if I'm not mistaken. Consider the superposition operator corresponding to $\operatorname{id}^p$, i.e. the operator $$T \colon L^p(E) \to L^1(E)$$ with $$(Tg)(x) = g(x)^p$$ for every $g \in L^p(E)$. A well-known result for superposition operators (see e.g. Theorem 1.2.1 in Progress In Nonlinear Differential Equations and Their Applications or Remark 2.5 in A Primer of Nonlinear Analysis) says that, because of its limited growth and pointwise continuity, $T$ is continuous. In other words, $f_n^p = Tf_n \to Tf = f^p$ in $L^1$.