I was reading proofs of the Riesz Theorem, and in most version there is such a line.
Let $\phi$ be an element of the dual of a Hilbert space, where the base field is denoted $F$.
Then if $\ker \phi$ is the kernel of $\phi$, we can choose an element $z$ that is perpendicular to $\ker \phi$, and $z\neq 0$.
So here I have assumed that $\phi$ is not 0 everywhere in the space so there is at least a one-dimensional subspace in the complement of the kernel. How am I sure however that there is an element in the complement that is perpendicular to the kernel, except the zero vector?
I also know that this is to show that the orthogonal complement of the kernel, which is a subspace, is not the set {0}.
I know you can decompose the Hilbert space by showing it is isomorphic to the direct sum of $F$ and $\ker\phi$. But can we not rely on this fact and still prove this statement?
Best Answer
The point is that $\phi$ is a continuous functional and so $\ker \phi = W$ is a closed subspace of $H$. Given a closed subspace $W$ of a Hilbert space $H$, you always have the direct sum decomposition $W \oplus W^{\perp} = H$. Thus, if $W \neq H$, you must have a non-zero vector in $W^{\perp}$.