quantum mechanics - Time derivative of the state vector as expressed in abstract Hilbert space vs. as a wavefunction

The Schrodinger equation in Hilbert space is expressed as : $$\frac{\partial}{\partial t} \psi(t) = \frac{-i}{\hbar}H\psi(t).$$

Here $\frac{\partial}{\partial t} \psi(t) \equiv \psi'(t) \equiv\lim \limits_{h \to 0} [\frac{\psi(t+h)-\psi(t)}{h}]$, and because $\psi(t)$ is a Hilbert space vector, the limit is defined using convergence via the norm. (In other words, for any $\epsilon>0$ there exists an $h_\epsilon>0$, such that $\big|\big|\psi'(t)-(\frac{\psi(t+h)-\psi(t)}{h})\big|\big| < \epsilon$ for all $|h|as a whole.

But in the wavefunction realization (for a single particle), the Schrodinger equation is expressed as

$$\frac{\partial}{\partial t} \psi(x;t) = \frac{-i}{\hbar}\left [ - \frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V \right ] \psi(x;t).$$

Here, however, $\frac{\partial}{\partial t} \psi(x;t)$ is a pointwise partial derivative with respect to $t$, and so the convergence depends only on each individual point $x$ of $\psi(x)$ separately.

If we take the abstract Hilbert space expression as definitive (axiomatically), then how can it be shown that the wavefunction realization actually expresses the same thing, given the different meanings of $\frac{\partial}{\partial t}$?