The function sequence $(\exp(f_n))$ converges uniformly if the function sequence $(f_n)$ converges uniformly.

Question

I am trying to prove that if the function sequence $(f_n)$ on $X \subset \mathbb{C}$ converges uniformly to $f: X \rightarrow \mathbb{C}$, and the function $\textrm{Re}(f)$ (the real values of $f$) is bounded, then the function sequence $(\exp(f_n))$ converges uniformly on $X$.

I really don't know how to start, i tried seeing $(\exp(f_n))$ as a sum, but i didn't know what to do there. I'll be so grateful with any clue to solve this problem. Thanks.

Answer 1

Write $$ |\exp(f_n(z)) - \exp(f(z))| = |\exp(f(z))| \cdot | \exp(f_n(z) - f(z)) - 1 | \\ = |\exp(\operatorname{Re}(f(z)) | \cdot | \exp(f_n(z) - f(z)) - 1 | \,. $$ Now use that the first factor is bounded, that $f_n(z) - f(z) \to 0$ uniformly in $X$, and that the exponential function is continuous.