I have some given maps $f_n:[0,T] \to H$ that are continuous maps into a Hilbert space $H$. For any $\epsilon > 0$, there exist numbers $T_n > 0$ such that $$\lVert f_n(t) \rVert \leq \epsilon|t| \quad \forall t \in [0,T_n].$$
We have that
- for each $n$, $f_n(t) \to 0$ as $t \to 0^+$
- for each $n$, $t^{-1}f_n(t) \to 0$ as $t \to 0^+$
- for each $t \in [0,T]$, $f_n(t) \to f(t)$ as $n \to \infty$
- for each $n$, $f_n$ is Lipschitz continuous wrt. $t$.
- $f$ also satisfies points 1, 2 above, and is also continuous
Is it possible that either $f_n(t) \to 0$ as $t \to 0^+$ uniformly in $n$ or $t^{-1}f_n(t) \to 0$ as $t \to 0^+$ uniformly in $n$?
I don't believe either follows. I tried a contradiction argument to prove it but was unsuccessful. But maybe I missed something?
Is it possible? Of course. As a trivial example each $f_n$ could be identically $0$. You are wondering if it is necessary that these conditions hold.
You can work with real functions if you consider $\phi_n(t) = \|f_n(t)\|$, and conversely given functions $\phi_n$ you can construct a counterexample by defining $f_n(t) = \phi_n(t)u$ for some unit vector $u$.
Consider functions $\phi_n$ defined by $$\phi_n(t) = \left\{ \begin{array}{cl} 0 & 0 \le t \le \frac 1n \\ n t - 1 & \frac 1n \le t \le \frac 2n \\ -nt + 3 & \frac 2n \le t \le \frac 3n \\ 0 & \frac 3n \le t \le T. \end{array} \right.$$ starting at an index $n$ satisfying $\frac 3n < T$ (or assume without loss of generality that $T > 3$, or modify the definition accordingly). Despite the odd appearance, these are just functions that vanish off the interval $(1/n,3/n)$, satisfy $\phi_n(2/n) = 1$, and are piecewise linear.
Each $\phi_n$ satisfies the stated conditions since $\phi_n$ vanishes in a one-sided neighborhood of $0$, and each $\phi_n$ has Lipschitz constant $n$. Moreover $\phi_n(t) \to 0$ for all $t$ so its limit is also a continuous function satisfying the stated conditions.
However, in any one-sided neighborhood of $0$ there is an index $n$ with $\phi_n(t) = 1$ for some $t$ in the neighborhood. Thus $\phi_n$ cannot converge uniformly to $0$, and obviously $\phi_n(t)/t$ cannot either.