Norm induced by Banach space and a Linear Transformation

Question

Let E be a normed vectorial space on $\mathbb{K}$ and F a vectorial space on $\mathbb{K}$ .
$L : E \rightarrow F$ a surjective linear operator such that $L^{-1}(0)$ is closed.

1. Prove that $$\|y\| = \inf\{\|u\| \mid u \in L^{-1}(y)\}$$ is a norm in F.
2. If E is Banach then F is Banach, with the norm defined in 1.

In question 1 the only problem that i have is in prooving that if $\|y\| = 0 \Longrightarrow y = 0$.

I understand that what you do by using linear operator is giving a norm to the space F using the norm in E, but for trying to prove $y = 0$ in all ways i tried, always need continuity of L. I think that the problem information in too weak for continuity and i am not sure but think that if $L^{-1}(y)$ is closed for all $y$ that could help. Is there a way of expand that condition that holds for $y = 0$ for all y in $F$?

About question 2, choosing a cauchy sequence in F, tried to prove that it converges using a sequence of pre-images of each element but, however, $L^{-1}$ could not be a function so in order to choose the right sequence in E i guess that if $L^{-1}(y)$ is closed, i could choose the element in pre-image such that its norm equals the infimun, but even if i could do that, have no idea about how to conclude.

Answer 1

For 1., you can consider the technique given by AlexL.

For 2., I would recommend using the following characterization of Banach spaces:

Let $X$ be a normed vector space. Then $X$ is a Banach space if and only if for every sequence $(x_n)$ in $X$, if $\sum_k\|x_k\|$ converges, then $\lim_{n\to\infty}\sum_{k=1}^nx_k$ converges in $X$.

Answer 2

1. Suppose $\parallel y \parallel =0$. Then there is a sequence $(u_n)$ of $L^{-1}(y)$ s.t. $\parallel u_n \parallel$ converges to $0 \in \mathbb{R}$, that is to say $u_n$ converges to $0_E$. But $L^{-1}(y)$ is closed (because if you choose $x$ in $L^{-1}(y)$, then $ L^{-1}(y)=x+L^{-1}(0)$). So $0_E \in L^{-1}(y)$ and $y=L(0_E)$.