Let $H$ a Hilbert space and $T \in L(H)$ a linear self-adjoint operator, I cannot find the theorem on the internet that tells me that $\ker(T)^{\bot} = \overline{\operatorname{Ran}(T^*)}$.
I just want to understand why I have to take the closure on the range of the adjoint.
Be warned that range of an operator is not always closed. For instance, consider the diagonal operator on $\ell ^2$ given by $T\left( x_1, x_2, \ldots, x_n, \ldots \right) = \left( x_1, \frac{x_2}{2}, \ldots, \frac{x_n}{n}, \ldots \right)$. Had the range of this operator been closed, then there would exist some $C>0$ such that $\lVert T x \rVert \ge C \lVert x \rVert$ for each $x \in \ell^2$ for this operator $T$ is injective (see here). Now, if we take $x=e_n$ for $n \in \mathbb N$, then we have $\frac{1}{n} \ge C$. Since $n$ was arbitrary, $C=0$ which is a contradiction.
As far as your question is concerned, given any subset $M$ of a Hilbert space $H$, $M^{\perp}$ is always closed as you can check. Consequently, for any bounded operator $T$ on a Hilbert space $H$, $(\ker T)^{\perp}$ must be closed, nevertheless, $\operatorname{ran} T^{*}$ might not be closed as we have seen. This suggests why we need to take the closure of $\operatorname{ran} T$ instead of just $\operatorname{ran} T^{*}$.
For a proof, see S. Axler, Measure, Integration and Real Analysis, Theorem 10.13.
