Difference of seminorms

Question

Given a Hilbert space $H$, take a continuous functional $f\in X^*$ and define a map $\|\cdot\|':H\to\mathbb R$ by $\|y\|^\prime = \|y\| - |f(y)|$. Can it happen that this is a Hilbert-space norm when $f$ is non-zero?

Answer 1

I think that $\|\cdot\|'$ is not a norm, but here is an easy way to at least show that it is not a Hilbert norm.

For better notation, let me write $\|y\|_0=\|y\|-|f(y)|$.

Let $z_0\in (\ker f)^\perp$ with $f(z_0)=1$. Then $H=\ker f+\mathbb C z_0$. So any $x\in H$ is of the form $x=x_0+\lambda z_0$, with $x_0\in\ker f$. Note that $\|x_0\|_0=\|x_0\|$. We have, using the Parallelogram Law, $$\tag1 \|x_0+\lambda z_0\|_0^2+\|x_0-\lambda z_0\|_0^2=2\|x_0\|_0^2+2|\lambda|^2\,\|z_0\|_0^2=2\|x_0\|^2+2|\lambda|^2(\|z_0\|-1)^2 $$ On the other hand, by definition, \begin{align}\tag2 \|x_0+\lambda z_0\|_0^2+\|x_0-\lambda z_0\|_0^2 &=(\|x_0+\lambda z_0\|-|\lambda|)^2+(\|x_0-\lambda z_0\|-|\lambda|)^2\\ \ \\ &=\|x_0+\lambda z_0\|^2+\|x_0-\lambda z_0\|^2+2|\lambda|^2-2|\lambda|(\|x_0+\lambda z_0\|+\|x_0-\lambda z_0\|)\\ \ \\ &=2\|x_0\|^2+2|\lambda|^2\,\|z_0\|^2+2|\lambda|^2-2|\lambda|(\|x_0+\lambda z_0\|+\|x_0-\lambda z_0\|). \end{align} Comparing $(1)$ and $(2)$ we get $$\tag3 4|\lambda|^2\|z_0\|=2|\lambda|(\|x_0+\lambda z_0\|+\|x_0-\lambda z_0\|). $$ The equality $(3)$ will hold for any $\lambda$. First, for any nonzero $\lambda$ we get $$\tag4 2|\lambda|\|z_0\|=\|x_0+\lambda z_0\|+\|x_0-\lambda z_0\|. $$ Now making $\lambda\to0$ we obtain $2\|x_0\|=0$, a contradiction since the above should work for any $x_0$. Or, we can see the above as a proof that $\ker f=0$, so $H=\mathbb C$.

In this latter case, when $\dim H=1$, the norm $\|\cdot\|_0$ is a norm, provided that $\|f\|< 1$.