Inverse of $\Phi(f) =\operatorname{sgn}(f)\lvert f\rvert^{1/p}$ is continuous from $L^p(E)$ to $L^1(E)$ I am working through Real Analysis, fourth edition, by Royden and Fitzpatrick. I ask your help with the last sentence, in emphasized font below, of Problem 48 of Chapter 7. I have included two lead-up problems and the other parts of Problem 48, all of which I have solved.



*Show that for $1 < p <\infty$ and any two numbers $a$ and $b$,
$$\Bigl\vert\operatorname{sgn}(a)\cdot\lvert a\rvert^{1/p} -\operatorname{sgn}(b)\cdot\lvert b\rvert^{1/p}\Bigr\rvert^p\le 2^p\cdot\lvert a - b\rvert.$$

*Show that for $1 < p <\infty$ and any two numbers $a$ and $b$,
$$\vert\operatorname{sgn}(a)\cdot\lvert a\rvert^p -\operatorname{sgn}(b)\cdot\lvert b\rvert^p\rvert\le p\cdot\lvert a - b\rvert(\lvert a\rvert +\lvert b\rvert)^{p-1}.$$

*(Mazur) Let $E$ be a measurable set and $1 < p <\infty$. For $f$ in $L^1(E)$, define the function $\Phi(f)$ on $E$ by
$$\Phi(f)(x) =\operatorname{sgn}(f(x))\lvert f(x)\rvert^{1/p}.$$
Show that $\Phi(f)$ belongs to $L^p(E)$. Moreover, use Problem 46 to show that
$$\lVert\Phi(f) -\Phi(g)\rVert_p^p\le 2^p\cdot\lVert f - g\rVert_1\text{ for all $f,g$ in }L^1(E).$$
From this conclude that $\Phi$ is a continuous mapping of $L^1(E)$ into $L^p(E)$ in the sense that if $\{f_n\}\to f$ in $L^1(E)$, then $\{\Phi(f_n)\}\to\Phi(f)$ in $L^p(E)$. Then show that $\Phi$ is one-to-one and its image is $L^p(E)$. Find a formula for the inverse mapping. Use the preceding problem to conclude that the inverse mapping  $\Phi^{-1}$ is a continuous mapping from $L^p(E)$ to $L^1(E)$.

Here is my incorrect solution:

To see that $\Phi^{-1}$ is a continuous mapping from $L^p(E)$ to $L^1(E)$, consider a sequence $\{h_n\}$ of functions such that $\{h_n\}\to h$ in $L^p(E)$. Then each $h_n$ and $h$ belong to $L^p(E)$ so that by the formula $\Phi^{-1}(h)(x) =\operatorname{sgn}(h(x))\lvert h(x)\rvert^p$ that I found solving a previous part of this problem, each $\Phi^{-1}(h_n)$ and $\Phi^{-1}(h)$ belong to $L^1(E)$. Furthermore,\begin{align}
&\lim_{n\to\infty}\int_E\big\lvert\Phi^{-1}(h) -\Phi^{-1}(h_n)\big\rvert\\
& =\lim_{n\to\infty}\int_E\bigl\vert\operatorname{sgn}(h)\lvert h\rvert^p -\operatorname{sgn}(h_n)\lvert h_n\rvert^p\bigr\vert\\
&\le p\lim_{n\to\infty}\int_E\lvert h - h_n\rvert(\lvert h\rvert +\lvert h_n\rvert)^{p-1} &\text{Problem 47}\\
&\le p\int_E\lim_{n\to\infty}\lvert h - h_n\rvert(\lvert h\rvert +\lvert h_n\rvert)^{p-1} = 0\end{align}
where we passed the limit under the integral sign using Lebesgue's general dominated convergence theorem (Section 4.4) with dominating sequence $\{(\lvert h\rvert +\lvert h_n\rvert)^p\}$.
Consequently, $\bigl\{\Phi^{-1}(h_n)\bigr\}\to\Phi^{-1}(h)$ in $L^1(E)$ by the sentence before the definition on page 145. Thus, $\Phi^{-1}$ is a continuous mapping from $L^p(E)$ to $L^1(E)$.

My solution has two mistakes: (1) In order to pass the limit under the integral sign, $\{h_n\}$ needs to converge pointwise a.e. to some function and (2) $\lim_{n\to\infty}\lvert h - h_n\rvert = 0$. It is not true in general that just because $\{h_n\}\to h$ in $L^p(E)$, then $\{h_n\}\to h$ pointwise a.e. on $E$.
I do know that a subsequence of $\{h_n\}$ converges to $h$ pointwise a.e. on $E$ (I read this page but do not think it applies here). I also know that $\lim\limits_{n\to\infty}\int_E\lvert h_n - h\rvert^p = 0$ which implies $\lim\limits_{n\to\infty}\int_E\lvert h_n\rvert^p =\int_E\lvert h\rvert^p$. I know that the linear spaces of simple functions in $L^p(E)$ and continuous functions on $E$ that vanish outside a bounded set are each dense in $L^p(E)$.
The way the problem is stated, I feel like the solution is a simple consequence of Problem 47.
 A: With credit belonging to Jose27, I am answering and accepting my question for the sake of completeness. I am hopeful that someone will post an answer better or as good as mine, in which case I will accept that answer over mine.
To see that $\Phi^{-1}$ is a continuous mapping from $L^p(E)$ to $L^1(E)$, consider a sequence $\{h_n\}$ of functions such that $\{h_n\}\to h$ in $L^p(E)$. Then each $h_n$ and $h$ belong to $L^p(E)$ so that by the formula $\Phi(h)^{-1}(x) =\operatorname{sgn}(h(x))\lvert h(x)\rvert^p$ that I found solving a previous part of this problem, each $\Phi^{-1}(h_n)$ and $\Phi^{-1}(h)$ belong to $L^1(E)$. Furthermore,\begin{align}
&\lim_{n\to\infty}\int_E\big\lvert\Phi^{-1}(h) -\Phi^{-1}(h_n)\big\rvert\\ & =\lim_{n\to\infty}\int_E\bigl\vert\operatorname{sgn}(h)\lvert h\rvert^p -\operatorname{sgn}(h_n)\lvert h_n\rvert^p\bigr\vert\\
&\le p\lim_{n\to\infty}\int_E\lvert h - h_n\rvert(\lvert h\rvert +\lvert h_n\rvert)^{p-1} &&\text{Problem 47}\\
&\le p\lim_{n\to\infty}\lVert h - h_n\rVert_p\cdot\bigl\Vert(\lvert h\rvert +\lvert h_n\rvert)^{p-1}\big\rVert_{p/(p-1)} &&\text{Theorem 1}\\
&\le p\cdot 0\cdot\lim_{n\to\infty}\big\lVert\lvert h\rvert +\lvert h_n\rvert\big\rVert_p^{p-1} = 0.\end{align}
In order to apply Hölder's inequality (the first part of Theorem 1), function $(\lvert h\rvert +\lvert h_n\rvert)^{p-1}$ must belong to $L^{p/(p-1)}(E)$, which is true by the second part of Theorem 1 because each $\lvert h_n\rvert$ and $\lvert h\rvert$ belong to $L^p(E)$, and $L^p(E),L^{p/(p-1)}(E)$ are linear spaces. Next, that $\lim\limits_{n\to\infty}\lVert h - h_n\rVert_p = 0$ is just our starting assumption $\{h_n\}\to h$ in $L^p(E)$. Finally, we are justified in setting the next-to-last expression to zero because $\lim\limits_{n\to\infty}\lVert\lvert h\rvert +\lvert h_n\rvert\rVert_p^{p-1}$ is bounded since, once again, each $\lvert h_n\rvert$ and $\lvert h\rvert$ belong to $L^p(E)$. Consequently, $\bigl\{\Phi^{-1}(h_n)\bigr\}\to\Phi^{-1}(h)$ in $L^1(E)$. Thus, $\Phi^{-1}$ is a continuous mapping from $L^p(E)$ to $L^1(E)$.
