A second proof of the fact that a linear functional $f$ on a normed vector space $\mathcal{X}$ is bounded iff $f^{-1}(0)$ is closed.

Question

I am working through Folland, not for homework, and have come across this problem (Ch 5 #17):

A linear functional $f$ on a normed vector space $\mathcal{X}$ is bounded iff $f^{-1}(0)$ is closed.

Note the "only if" direction is trivial. Folland suggests that we use the following fact in our proof: for all $\epsilon>0$ there exists an $x\in\mathcal{X}$ such that $\|x\|=1$ and $\|x+\mathcal{M}\|\ge 1-\epsilon$, where $\mathcal{M}$ is a proper closed subspace of $\mathcal{X}$.

I can prove both the hint, and the exercise, but not the latter using the former! My proof is to suppose (toward a contradiction), that there existed some sequence $\|x_n\|=1$ such that $f(x_n)>n$ for each $n\in\mathbb{N}$, and then consider the sequence $\{x-f(x)\frac{x_n}{f(x_n)}\} $ and observe that this is in the kernel of $f$ for each $n$ and approaches $x$ in the limit, from which the statement should follow since $x$ is arbitrary.

Can anyone tell me how to use the hint? I'm guessing the closed subspace we consider is the kernel of $f$, but it's not clear how to go from there.

Any help would be appreciated.

Answer 1

The hint gives you an $x\in X$, with $\|x\|=1$, and such that $x\not\in\ker f$ (because this is equivalent to $\|x+\ker f\|\ne0$). Now, since $f(x)\ne0$ we can write, for any $y\in X$, $$ y=\frac{f(y)}{f(x)}\,x+\left(y-\frac{f(y)}{f(x)}\,x\right), $$ where the term in brackets belongs to $\ker f$. So $X=\mathbb{C}x\oplus\ker f$. If we write $k=|f(x)|$ we get, for any $\lambda x+m\in\mathbb{C}x\oplus\ker f$, $$ |f(\lambda x+m)|=|f(\lambda x)|=|\lambda|\,|f(x)|=k\,|\lambda|. $$ From the hint, we also have $$ \|\lambda x+m\|=|\lambda|\,\|x+\lambda^{-1}\,m\|\geq|\lambda|\,(1-\varepsilon). $$ Joining the two estimates we get $$ |f(\lambda x+m)|=k\,|\lambda|\leq\,\frac{k}{1-\varepsilon}\,\|\lambda x + m\|. $$ So $f$ is bounded.

Answer 2

The linear map $f$ factor through the quotient $F: X/\ker(f)\to \mathbb R$ and it is continuous since is a linear map between finite dimensional subspaces. On the other hand, the quotient map $\pi:X/\ker(f)\to \mathbb R$ is continuous as $\ker(f)$ is closed, hence $f=F\circ \pi$ so it is a composition of continuous functions.

Answer 3

I think you original argument is much more instructive and easily generalizable. We can prove something stronger by tinkering with it and I will do this to complement the other excelent answer by Argerami:

If $f^{-1}(\alpha)$ is closed and non empty, $f$ is bounded.

Modification of your Proof: Take $x_o\in f^{-1}(\alpha)$ and suppose by way of contradiction that $f$ is not bounded in the circle. Take $y_j$ with $\lVert y_j\rVert=1$ and $|f(y_j)|\xrightarrow{j\rightarrow \infty} \infty$. In the norm of our vector space:

$$a_j=x-\frac{(f(x)-f(x_o))y_j}{f(y_j)}\rightarrow x$$

Because each $a_j\in f^{-1}(\alpha)$ and $f^{-1}(\alpha)$ is closed, we need $f(x)= \alpha$, but $x$ is arbitrary, so $f(y_j)=\alpha$ in contradiction to the fact $|f(y_j)|\xrightarrow{j\rightarrow \infty} \infty$. $\square$

Alternative more Geometric Proof: If $f$ is identically equal to $\alpha$, we need $\alpha=0$ as $f(0)=0$. In this case $f$ is trivially bounded.

If $f$ is not identically $\alpha$, take a certain element $x_o\in f^{-1}(\alpha)^C$ and define a ball $B_{x_o}(R)\subseteq f^{-1}(\alpha)^C$. Suppose without loss of generality that $f(x_o)>\alpha$.

If there were a certain point $x_1$ where $f(x_1)<\alpha$ in this ball, we would have:

$$x=x_1+\frac{x_o-x_1}{f(x_o)-f(x_1)}(\alpha-f(x_1))\in f^{-1}(\alpha)$$

$$ \lVert x-x_o\rVert=\Vert x_1-x_o\rVert\left|1+\frac{\alpha-f(x_1)}{f(x_1)-f(x_o)}\right|=\Vert x_1-x_o\rVert\left|\frac{\alpha-f(x_o)}{f(x_1)-f(x_o)}\right|\leq \Vert x_1-x_o\rVert\leq R$$

But this is absurd as $f^{-1}(\alpha)\cap B_{x_o}(R)=\emptyset$. Therefore, we need that for all $x$ with $\lVert x \rVert=1$:

$$f(x_o+Rx)> \alpha \Rightarrow f(-x) <\frac{f(x_o)-\alpha}{R}=K$$

In particular, $f(x)=f(--x)<K$, so in fact $|f(x)|\leq K$. For an arbitrary $y$, $y/\lVert y \rVert$ is unitary so:

$$|f(y)|\leq \frac{f(x_o)-\alpha}{R} \lVert y\rVert \quad \quad \square$$