Question: If $T$ is a densely defined operator from a subspace of a Hilbert space $H$ to a Hilbert space $K$, how to prove that $\mbox{Ker}(T)=\mbox{Ker}(T^\*T)$?

Answer: 
The fact that $T$ is only densely defined does not change the proof of the bounded case $T\in B(H)$. Indeed, if $x$ belongs to $\mbox{Ker} T$, then $Tx=0$ belongs to the domain of $T^\*$. Now if $x$ belongs to the domain of $T^\*T$, a fortiori it belongs to the domain of $T$. So the following holds like in the bounded case.

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> $$Tx=0\;\Rightarrow \; T^\*Tx=0\;\Rightarrow\; (x,T^\*Tx)=\|Tx\|^2=0\;\Rightarrow\; Tx=0$$
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One direction is trivial. For the other, consider an element in the kernel of $T^\*T$, and the definition of the adjoint, and of the properties of the inner product.