Negation of uniform convergence

Question

Suppose $f_{n}$ is a sequence of functions which does not convergence uniformly to $f$. Does this mean that there exists an $\varepsilon_{0} > 0$, an $x_{0}$, and a sequence of integers $n_{k} \rightarrow \infty$ such that $|f_{n_{k}}(x_{0}) - f(x_{0})| \geq \varepsilon_{0}$?

Answer 1

If you know how to negate logical formulas with quantifiers, you can do this more or less mechanically.

Definition of uniform convergence can be written like this: $$(\forall \varepsilon>0) (\exists n_0) (\forall x\in S) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon) $$ Negation of uniform convergence $$(\exists \varepsilon>0) (\forall n_0) (\exists x\in S) (\exists n>n_0) (|f_n(x)-f(x)|\ge\varepsilon) $$

Pointwise convergence is defined as follows $$(\forall \varepsilon>0) (\forall x\in S) (\exists n_0) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon) $$ negation $$(\exists \varepsilon>0) (\exists x\in S) (\forall n_0) (\exists n>n_0) (|f_n(x)-f(x)|\ge\varepsilon)$$ If you look closely at the negation of pointwise convergence, it is equivalent to condition from your question. Indeed, we have existence of $\varepsilon>0$ and existence of a point, which we many denote $x_0$, such that $|f_n(x_0)-f(x_0)|\ge\varepsilon$ happens for infinitely many $n$'s. So any function which converges pointwise but not uniformly is a counterexample to the claim in your post.