Convergence of Exponential in $L^p$ Let $(X, \mu)$ be a finite measure space and let $f_n$ be a sequence of functions in $L^p(X, \mu)$, $1 \leq p < \infty$ such that $f_n \longrightarrow f$ in $L^p$ as $n$ tends to infinity. Suppose that $e^{f_n}, e^f \in L^p(X, \mu)$ with 
$$ \|e^f_n\|_{L^p}, \|e^f\|_{L^p} \leq C$$
for some constant $C$ and all $n$.
Can I conclude that $e^{f_n} \longrightarrow e^f$ in $L^p(X, \mu)$?
\Edit: So by the example below one cannot conclude convergence in $L^p$, but can one conclude convergence in $L^q$ for $q<p$? Namely, by the reference in the comment, one can at least conclude the existence of a subsequence that does this.
 A: For $p < +\infty$, you cannot conclude the $L^p$ convergence of the exponentiated sequence. Let's consider $X = (0,1]$, and $\mu$ the Lebesgue measure. Fix $p \in [1,+\infty)$, and let $g(x) = x^{-\frac{1}{2p}}$. For $0 \leqslant a < b \leqslant 1$ we have
$$\lVert g\cdot \chi_{(a,b]}\rVert_{L^p}^p = \int_a^b \frac{dx}{\sqrt{x}} = 2(\sqrt{b} - \sqrt{a}) \leqslant 2\sqrt{b}.$$
So for any sequence of intervals $(a_n,b_n]$ with $b_n \to 0$, we have $f_n \to 0$ in $L^p(X,\mu)$ where $f_n = g\cdot \chi_{(a_n,b_n]}$.
But for $e^{f_n}$, we have (dropping the indices on the interval bounds)
\begin{align}
\lVert e^{f_n}\rVert_{L^p}^p &= 1 - (b-a) + \int_{a}^{b} \exp \biggl(p\cdot x^{-\frac{1}{2p}}\biggr)\,dx\\
&= 1 - (b-a) + \int_{1/b}^{1/a} \frac{\exp \bigl(p \sqrt[2p]{y}\bigr)}{y^2}\,dy\\
&= 1 - (b-a) + 2p\int_{b^{-\frac{1}{2p}}}^{a^{-\frac{1}{2p}}} \frac{e^{pt}}{t^{2p+1}}\,dt,
\end{align}
and since
$$\int_c^\infty \frac{e^{pt}}{t^{2p+1}}\,dt = +\infty$$
for every $c > 0$, we can for every $b_n > 0$ choose $a_n \in (0,b_n)$ so that $\lVert e^{f_n}\rVert_{L^p} = 2$, so that $e^{f_n} \nrightarrow e^f = 1$ in $L^p$ although the sequence is uniformly bounded (in the $L^p$-norm) and converges pointwise.
For $1 < p < \infty$, we can at least conclude weak convergence $e^{f_n} \rightharpoonup e^f$. Since every subsequence of $(e^{f_n})$ has a further subsequence that converges pointwise almost everywhere to $e^f$, the only possible weak limit of subsequences of $e^{f_n}$ is $e^f$.
For suppose we have a subsequence such that $e^{f_{n_k}}$ converges weakly to $g$. By the $L^p$ convergence $f_n \to f$, we can extract a further subsequence so that $f_{n_{k_m}}$ converges pointwise (a.e.) to $f$. To simplify typography, let's assume that $f_n\to f$ pointwise a.e. and $e^{f_n} \to g$ weakly. Suppose $g \neq e^f$. Then without loss of generality
$$A = \bigl\{x : \operatorname{Re} e^{f(x)} > \operatorname{Re} g(x) \bigr\}$$
has positive measure. There is a measurable $B\subset A$ with $\mu(B) > 0$ and a constant $K < +\infty$ such that $\lvert f(x)\rvert \leqslant K$ and $\lvert g(x)\rvert \leqslant e^K$ on $B$. By Egorov's theorem, there is a measurable $D\subset B$ with $\mu(D) > \frac{1}{2} \mu(B)$ such that $f_n \to f$ uniformly on $D$. Without loss of generality, we may assume that $\lvert f_n(x)\rvert \leqslant K+1$ on $D$ (drop finitely many terms from the sequence). The continuity of the exponential function then shows that $e^{f_n} \to e^f$ uniformly on $D$. Then $\chi_D \in L^q(X,\mu) \cong L^p(X,\mu)^{\ast}$, and
\begin{align}
\operatorname{Re} \int_X g\chi_D\,d\mu &= \operatorname{Re} \int_D g\,d\mu\tag{weak convergence}\\
&= \operatorname{Re} \lim_{n\to\infty} \int_D e^{f_n}\,d\mu\\
&= \operatorname{Re} \int_D e^f\,d\mu\tag{uniform convergence}\\
&= \int_D \operatorname{Re} e^f\,d\mu\\
&> \int_D \operatorname{Re} g\,d\mu\\
&= \operatorname{Re} \int_D g\,d\mu,
\end{align}
which is absurd. Hence $g \neq e^f$ is impossible.
Since $L^p(X,\mu)$ is reflexive for $1 < p < +\infty$, and $(e^{f_n})$ is bounded in $L^p$, every subsequence has a weakly convergent subsequence. It follows that the whole sequence converges weakly to $e^f$.
