Let $f_k(x) = \frac{e^{-kx}}{k^2}$ where $x \in [0, 1]$, $k \in \mathbb{N}$. Is this sequence equicontinuous? What if $x \in \mathbb{R}$?
The definition I have been given is: Let $F$ be a subset of $\mathscr{F}(X;Y)$, functions from metric space $(X, d_X)$ to $(Y, d_Y)$. We say $F$ is equicontinuous if for all $\epsilon > 0$ there exists $\delta > 0$ such that whenever $d_X(x, y) < \delta$, for $x, y \in X$, we have $d_Y(f(x), f(y)) < \epsilon$ for all $f \in \mathscr{F}$.
I have tried arranging $|f_k(x) - f_k(y)| < \epsilon$ to something contianing $|x-y|$ but I am not yet successful. Any hints would be great.
Hint:
$$|f_k(x) - f_k(y)| = \left|f_k'(\xi)(x-y) \right|= \frac{ke^{-k\xi}}{k^2}|x-y|, \quad \xi \in (x,y)$$