Let $(H;(\cdot|\cdot))$ be separable Hilbert space where $\dim$ $H=\infty$. We define norm $\|x\|=\sqrt{\langle x|x\rangle}$ and linear operator $A:H\to H$. Show that:
a) functional $f(x):=\langle Ax|a \rangle \in \mathbb{C}$ is unbounded and $\ker f$ is dense in $H$, when vector $a\notin \operatorname{dom} A^{*}$, $ A^{*}:H\to H$.
b) show that $\|A\|=\infty$.
I would ask for a draft proof or a solution because I don't know where to start.
If $f$ is bounded then (by Riesz Theoem) there exists $y$ such that $\langle Ax , a \rangle =\langle x , y \rangle$ for all $x$. By definition this means $a \in dom(A^{*})$ and $A^{*}a=y$. This is a contradiction. Hence $f$ is unbounded.
Now $ker f$ has co-dimension $1$. (This is true for any non-zero linear functional). If $ker (f)$ is a proper subspace of $H$ then we would have the struct inclusions $\ker (f) \subset \overline {\ker (f)} \subset H$ but this contradicts the fact that co-dimension is $1$. Hence $ker (f)$ is dense. Of course $\|A\|=\infty$ becasue $A$ is unbounded.