This question is a continuation of my question "Exponential and convergence in $L^2$" posted above:
Let $(f_k)$ be a sequence of elements of $L^\infty(\Omega)$, which converge in $L^2(\Omega)$ to $f\in L^2(\Omega)$. Where $\Omega $ is an open bounded subset of $R^n$. Assume that $e^{f_k}\in L^2(\Omega), \; k\ge 1$ and $e^{f} \in L^2(\Omega)$.
Is it true that : $e^{f_k} $ tends to $e^f$ in $L^2(\Omega)$?
Thanks again
Define $\Omega=(0,1)$, $$ f_n(x) = \chi_{[0,1/n]} n^{1/4}. $$ Then $f_n\to0$ in $L^2(\Omega)$. And $$ \|e^{f_n}\|_{L^2(\Omega)^2}= n^{-1} e^{2n^{1/4}} + 1 - \frac 1n\to \infty, $$ so it holds $e^{f_n}\in L^2(\Omega)$, $e^f\in L^2(\Omega)$, but $(e^{f_n})$ cannot converge in $L^2$ as it is unbounded.
As it turns out, even uniformly boundedness of $(e^{f_n})$ in $L^2(\Omega)$ is not enough: with $$ f_n(x) = \chi_{(0,e^{-n})}(x) \frac n2, $$ it holds $$ \|f_n\|_{L^2(\Omega)}^2 = e^{-n} \frac {n^2}4 \to 0, $$ so $f_n \to f$ with $f=0$, and $$ \|e^{f_n}\|_{L^2(\Omega)}^2 = e^{-n} e^n + (1-e^{-n}) \to 2 \ne 1 =\|e^{f}\|_{L^2(\Omega)}^2 . $$ Hence $(e^{f_n})$ is uniformly bounded in $L^2(\Omega)$ but cannot converge to $e^f$ in $L^2(\Omega)$.
Under the additional assumption that $(f_n)$ is uniformly bounded in $L^\infty(\Omega)$, you can prove the desired convergence by Lebesgue dominated convergence theorem.