# Why is it true that the multiplication operator in a reproducing kernel Hilbert space is always continuous?

In my functional analysis I was met with this seemingly trivial theorem on RKHS

If $\mathbb{H}$ is a reproducing Kernel Hilbert Space and we have a multiplier $\phi$ meaning it satisfies $\forall f \in \mathbb{H} : \phi f \in \mathbb{H}$ and it is claimed the multiplication operator on $\mathbb{H}$ is continuous.

One source says this follows from the closed graph theorem which I cannot sseem to so easily apply here, and another source said this is namely due to the reproducing kernel property with the inequality:

$\langle \phi f, K_x \rangle$ = $\phi(x) \langle f, K_x \rangle$

This does not show me how convergence in norm of the arguments $f_n$ leads to convergence in norm of $\phi f_n$ so this is where I am stuck and truly need help.

The closed graph theorem indeed suggests itself: Suppose that $f_n\to f$ and $\phi f_n\to g$ in norm. We must show that then $g=\phi f$. This follows from $$g(x)=\langle K_x, g\rangle = \lim\, \langle K_x, \phi f_n\rangle =\phi(x) \lim\, \langle K_x, f_n\rangle =\phi(x)f(x) .$$