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 $R^n$. Is it true that : $e^{f_k} $ tends to $e^f$ in $L^2(\Omega)$? ($e$ is the exponential function).
I tested the inequality : $|e^x-e^y|\le \alpha |x-y|,\; \alpha>0$ in any bounded set of $R$, but I have no information about the boundednessof $f(x),\; x\in\Omega$ and $f_k(x),\; k\ge 1, \; x\in \Omega.$
Thanks
No, just consider e.g. $\Omega = (0,1)$, $$f_k(x) := x^{-1/4} \cdot 1_{(k^{-1},1)}(x)$$ and $$f(x) := x^{-1/4}.$$ Then $f_n \to f$ in $L^2$ (e.g. by the monotone convergence theorem), $\|f_k\|_{\infty} = k^{1/4}<\infty$, but $$e^{f(x)} = \exp \left(x^{-1/4} \right) \notin L^2.$$ This means in particular that $e^{f_k}$ cannot converge in $L^2$ to $e^f$.