why if $f$ is a bounded linear functional then ker $f$ is a closed subspace of the hilbert space $H$? The book said: "if $f$ is a bounded linear functional then ker $f$ is a closed subspace of the Hilbert space $H$."
But I do not know why, could anyone clarify this for me please? 
 A: 
Hint: Let $(X,d_1)$ and $(Y,d_2)$ be metric spaces. The map $f:X \to Y$ is continuous $\iff$ $f^{-1}(Z)\subset X $ is closed for each closed subset $ Z \subset Y$.

A: Let the field of scalars be denoted by $\Bbb K$, where $\Bbb K = \Bbb R$ or $\Bbb K = \Bbb C$.  Then taking $f:H \to \Bbb K$, since $f$ is bounded we have a positive real constant $\mu$ such that, for all $x \in H$,
$\vert f(x) \vert \le \mu \Vert x \Vert, \tag 1$
where $\Vert x \Vert = \langle x, x \rangle^{1/2}$ is the usual norm on $H$, induced by the inner product $\langle \cdot, \cdot \rangle$ on $H$.
Suppose $x_n \in H$ is a sequence of vectors in $\ker f$:
$f(x_n) = 0, \; \forall n \in \Bbb Z_+, \tag 2$
where $\Bbb Z_+ = \{m \in \Bbb Z \mid m > 0 \}$, and suppose that $x_n$ is Cauchy in $H$; this means that for every real $\epsilon > 0$ we can find $N \in \Bbb Z_+$ such that, for $i, j > N$,
$\Vert x_i - x_j \Vert < \epsilon; \tag 3$
since $H$ is Hilbert, the sequence $x_n$ converges to some $x \in H$; that is, for any positive $\epsilon$, taking $N$ suffiently large, we have,
$\Vert x - x_i \Vert < \epsilon, \tag 4$
provided $i > N$.  Then
$\vert f(x) \vert = \vert f(x) - f(x_i) \vert = \vert f(x - x_i) \vert \le \mu \Vert x - x_i \Vert < \mu \epsilon, \tag 5$
where we have used the linearity of $f$ to assert $f(x) - f(x_i) = f(x - x_i)$, and the assumption that $f(x_i) = 0$ to affirm $f(x) = f(x) - f(x_i)$.  Now since $\epsilon$ is an arbitrary positive real, (5) forces
$f(x) = 0; \tag 6$
thus $x \in \ker f$ and hence $\ker f$ is closed in $H$.
The above demonstration is essentially the same as arguing that the continuity (that is, the boundedness) of $f$ implies that the inverse image of the closed set $\{ 0 \} \subset \Bbb K$ is closed in $H$.
Note: $H$ need merely be Banach, not necessarily Hilbert, for the above argument to fly; in that case $\Vert \cdot \Vert$ is not generally induced by in inner product, but all else said above holds.  End of Note.
