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 $f(x)$?

  • 1
    $\begingroup$ What is $||\cdot||$? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.