Let $(X,\rho)$ be a compact metric space, we say a function on $X$ is uniformly $\beta$-continuous if, for every $\epsilon > 0$, there exists $\delta > 0$ such that if $\rho(x,y) < \delta$, $|f(x) - f(y)| < \epsilon + \beta$. Suppose $f_n \to f$ uniformly, and $f_n$ is $\beta_n$-continuous for $\beta = 1/n$. Is $f$ continuous? Prove or disprove.
I think the statement is true, and I tried to prove it by letting $\epsilon > 0$, we can find $\delta > 0$ such that if $|x-y| < \delta$, $|f_n(x) - f_n(y)| < \epsilon/2 + 1/n$. Now, $f_n \to f$ uniformly, so for any $\eta > 0$, there exists $N$ such that $|f_n(x) - f(x)| < \eta$ for all $n \ge N$, and all $x \in X$. Pick $\eta = \epsilon/2$. Then, if $|x-y| < \delta$, $$|f_n(x) - f(x)| + |f_n(y) - f(y)| < \epsilon,$$ at this point I've gotten stuck, because my inequalities all seem to be going the wrong way for me to get any nice results. I'm wondering if I am just trying to approach this problem the complete wrong way, and if there is something about the compact metric space (maybe involving Ascoli-Arzela, or something?) that I would need to use to get a nice result. Thanks!
Your inequalities have already gotten you very close to the proof.
For any $x$ and we have that $|f(x)-f(y)|=|f(x)-f_n(x) +f_n(x)-f_n(y)+f_n(y)-f(y)|$, so $$|f(x)-f(y)|\le |f(x)-f_n(x)| +|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$$ Since $f_n$ tends to $f$ uniformly, the first and last absolute values in the sum tend to zero as $n\rightarrow \infty$ regardless of x and y. Now given $\epsilon$, pick n such that $\frac 1n<\frac {\epsilon}{2}$, and $|f(x)-f_n(x)|<\epsilon$ for any x. Then pick $\delta$ such that if $p(x,y)<\delta$ then $|f_n(x)-f_n(y)|<\frac {\epsilon}{2}+ \frac 1n$. Then for this $n$ and $y$ we have $$|f(x)-f(y)|\le|f(x)-f_n(x)| +|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\le \epsilon +\frac {\epsilon}{2}+\frac 1n +\epsilon<3\epsilon $$ So f is continuous.