# uniformly convergence

suppose $$f_n(x)$$ is a sequence of complex functions which converges uniformly to $$f(x)$$,$$g_n(x)$$ is a sequence of complex functions such that $$|f_n(x)-g_n(x)|\to 0$$.Can we conclude that $$g_n(x)$$ is uniformly convergent to $$f(x)$$?

• What is $||\cdot||$? – Guacho Perez Nov 30 '18 at 0:42

Unless $$\lvert{f_n(x) - g_n(x)}\rvert \rightarrow 0$$ is uniform, no. For instance, take $$f_n(x) \equiv 0$$ and $$f \equiv 0$$. Then $$f_n \rightarrow f$$ uniformly on $$(0, 1) \subset \mathbb{R} \subset \mathbb{C}$$. Take $$g_n(x) = x^n$$. Then $$\lvert{g_n(x)\rvert} \rightarrow 0$$ for all $$x \in (0, 1)$$ pointwise but not uniformly.
If $$\lvert{f_n(x) - g_n(x)\rvert} \rightarrow 0$$ uniformly, then your statement is true as $$\lvert{g_n(x) - f(x)\rvert} \leq \lvert{g_n(x) - f_n(x)\rvert} + \lvert{f_n(x) - f(x)}\rvert$$ and you can bound both terms uniformly by assumption.