How can one prove that if $f$ is linear with a closed kernel, then $f$ is continuous

Question

Let $(V,\|.\|)$ be a normed vector space, $ f:V\to \mathbb{R} $ be linear and $ \ker(f) $ closed. Then $f$ is continuous. How does one prove this?

My idea was to distinguish between two cases.

Because $ \ker(f) $ is closed all convergent sequences $ (v_n)_{n\in \mathbb{N}}\subseteq \ker(f) $ have the limit in $ \ker(f) $ which means $ \lim\limits_{n\to \infty} v_n=v\in \ker(f) $. So it applies $ \lim\limits_{n\to \infty} f(v_n)=0=f(v) $.

Now let $ v\in V\setminus \ker(f) $. Because $ \ker(f) $ is closed then $ V\setminus \ker(f) $ is open. With that I tried to show the existence of a constant $c>0$ such that $ 0<|f(v)|\leq c\cdot \|v\| $ for all $ v\in V\setminus \ker(f) $ but I don't know how to get this constant $ c>0 $. I only have this $$ 0<|f(v)|=\left | f\left(\|v\|\cdot \frac{v}{\|v\|}\right)\right |\stackrel{f\text{ is linear}}{=}\|v\|\cdot \underbrace{\left | f\left(\frac{v}{\|v\|}\right)\right |}_{???} $$

Both subcases would imply the continuity of $ f $.

Answer 1

If $f=0$, we’re done. So assume there is a $y$ such that $f(y)=1$. If $f$ is not continuous, there is a sequence $x_n \rightarrow 0$ such that $f(x_n)=1$ (*). Then $x_n-y \rightarrow -y$ is a sequence in the kernel of $f$, so $f(-y)=0$. We get a contradiction.

Proof of (*): let $z_n$ be a sequence converging to zero such that $f(z_n)$ does not go to zero. Up to extracting, flipping signs, and renormalizing, we can assume $f(z_n) \geq 1$. Then put $x_n=\frac{z_n}{f(z_n)}$.