Convergence stronger than uniform convergence

Question

Is there any kind of convergence stronger than uniform convergence?

As for example $f_n(x)\underset{n\rightarrow+\infty}{\longrightarrow}f(x)$ with a 'form' of convergence that implies uniform convergence.

And is there a type of convergence which is the last type, so to say, before stating that two things are the same? Or we can always make an stronger convergence after another without implying equivalence?

Thank you.

Answer 1

If you consider differentiable functions from $[0,1]$ to $\mathbb{R}$, you could use some kind of convergence of derivatives, e.g., $\int_0^1|f'_n - f'| \, \mathrm{d}x \to 0$.

Answer 2

Here is one, with series. Let $f_n=\sum_{k=1}^ng_k$ with $g_k:X\longrightarrow E$, where $X$ is any set and $E$ is a Banach space. We say that $f_n$ converges normally if $$ \sum_{k=1}^{+\infty}\sup_{x\in X}\|g_k(x)\|<\infty. $$ This implies the uniform convergence of $f_n$ on $X$. The converse is false. For an example of a uniformly not normally convergent series, take $$ \sum_{n\geq 1}\frac{1}{n}1_{\{n\}} $$ on $\mathbb{R}$.