Uniform Convergence and Sup

Question

Let $X \subset \mathbb{R}^d$ be open and let $(f_n)_{n \in \mathbb{N}} : X \to \mathbb{C}$ be a sequence of complex-valued functions. Suppose $f_n \to f$ uniformly on $X$. Fix $a \in X$. I would like to prove $$ \sup_{x \in X} |f_n(x)-f_n(a)| \to \sup_{x \in X} |f(x)-f(a)|. $$ I tried to rewrite $$ \left| \sup_{x \in X} |f_n(x)-f_n(a)| - \sup_{x \in X} |f(x)-f(a)| \right| $$ using only one supremum. I thought in general for $A,B \subset \mathbb{R}$ one has $$ |\sup A - \sup B| \leq |\sup (A-B)|, $$ but I don't see how this would help after getting $$ \sup_{x,y \in X} |f_n(x) - f_n(a)| - |f(y) - f(a)|. $$

Answer 1

Here is a sketch: $$ |f_n(x) -f_n(a)| = |f_n(x) -f(x) - f_n(a) + f(a) +f(x) - f(a)|$$ $$ \leq |f_n(x) -f(x)| + |f_n(a) - f(a)| +|f(x) - f(a)|$$

Now take $\sup_{x \in X}$ on both sides, noting that supremum is subadditive. Then take limits to get your result.

Answer 2

Assume that $f_n$ is a bounded function for any $n$. Denote $g_n(x)=f_n(x)-f_n(a), g(x)=f(x)-f(a)$. Since $f_n$ converges uniformly to $f$ one gets by the triangle inequality that also $g_n$ converges uniformly to $g$ in particular $||g_n||_{\infty}$ converges to $||g||_{\infty}$.