Let $\mathcal{H}$ be a Hilbert space, $L\in \mathcal{H}^*$ be a bounded continuous linear functional. Suppose that $\ker\:L \neq \mathcal{H}$.
We know that the dimension of $\ker \:L$ is $1$, i.e. there exists $x(L) \in \mathcal{H}$ with $\|x(L)\| = 1$ such that for all $y \in \ker\: L$, we can write $y = k x$ for some $k\in \mathbb{R}$.
My question is, is the function defined by $f: \mathcal{H}^* \to \mathcal{H}, f(L) \mapsto x(L)$ continuous? That is to say, if two linear functionals are 'close', are their kernels also 'close'? I can't either prove this is correct or give a counter-example.
(PS: this question is the same as asking: In Riesz representation theorem, is the representing element continuous to the linear functional?)
Your function $\ x(L)\ $ is not well-defined, because if $\ \phi(L)=e^{i\psi(L)}\ $, where $\ \psi\ $ is any real-valued function on $\ \mathcal{H}^*\ $, then $ \psi(L)x(L)\ $ will satisfy all the the same conditions you've given to specify $\ x(L)\ $. It's certainly possible to choose $\ \psi\ $ to make $ \psi(L)x(L)\ $ discontinuous.
Your question is not "the same as asking" whether the representing element in the Riesz representation theorem is a continuous function of the represented functional. That representative, $ x_L\ $, say, is unique, and satisfies the stronger condition, $\ L(y)= \langle x_L, y\rangle\ $ for all $\ y\in \mathcal{H}\ $. As Mindlack has already pointed out, the function $\ L\mapsto x_L\ $ is a linear isometry, and hence continuous.