A new norm defined on X

Question

I am currently reading a book which adresses topics in Banach space theory and while i was reading i came across the following definition:

Suppose that $X$ is a normed space and $Y$ is a subspace of $X^*$.Let us consider a new norm on $X$ defined by $$\left\lVert{x}\right\rVert_Y=\sup\{|y^*(x)| : y^* \in Y \text{ , } \lVert{y^*}\rVert=1\}$$

And i was wondering why the above equation defines a norm in $X$ since i am not able to justify the implication $$\lVert{x}\rVert_Y=0 \implies x=0$$

Thanks in advance !

Edit: Actually i will post the whole definition so it could be more clear

Definition :

Suppose that $X$ is a normed space and $Y$ is a subspace of $X^*$.Let us consider a new norm on $X$ defined by $$\left\lVert{x}\right\rVert_Y=\sup\{|y^*(x)| : y^* \in Y \text{ , } > \lVert{y^*}\rVert=1\}$$If there is a constant $0<c\leq 1$ such that for all $x \in X$ we have $$c\lVert{x}\rVert\leq \lVert{x}\rVert_Y\leq \lVert{x}\rVert$$then $Y$ is said to be a c-norming subspace of $X^*$.

So i guess the author means that $\lVert{x}\rvert_Y$ is a norm when $Y$ is a c-norming subspace of $X^*$?

Answer 1

It's indeed not true, think about finite dimensional $X$ or $Y=0$.

It's true, however, if $Y=X^*$ or if $Y$ is dense in $X^*$.
Without this assumption we can only state that it's a norm on the quotient space $X/(\bigcap_{f\in Y} \ker f) $.

To your edit: yes, it perfectly makes sense if $c\Vert x\Vert\le\Vert x\Vert_Y$ with a fixed $c>0$.

Answer 2

That implication need not be true in general. If you want to insist $Y \neq \{0\}$ try considering the case where $Y = \operatorname{span}(\phi)$ is one dimensional, for $\|\phi\| =1$ (and $\dim X > 1$). Then $\|x\|_Y = |\phi(x)|$. But $\phi$ must have non-trivial kernel so $\| \cdot \|_Y$ is not a norm in general.

In response to the edit: What's going on here is that in general $\| \cdot \|_Y$ defines a seminorm on $X$. However, whenever a seminorm is equivalent to a norm, as is the case if $Y$ is a $c$-norming subspace of $X^*$, then that seminorm is automatically actually a norm since $\|x\|_Y = 0 \implies \|x\| = 0$.