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I have a small issue in understanding the very last part of the proposition 1.5 in Brezis`

The statement of the proposition is:

The hyperplane $H = [f = \alpha]$ is closed if and only if $f$ is continuous.

Here $f$ is a linear functional defined on a normed vector space $E$ and

$$ H = [f = \alpha] = \left\{ x \in E : f(x) = \alpha \right\} $$

Firstly it does state that $f$ continuous $\Rightarrow H$ is closed is trivial. To me it isn't, so I want to prove it... the set $H = f^{-1}(\left\{ \alpha \right\})$, and ${\alpha}$ is a closed set, and since $f$ is continuous the preimage of $\left\{ \alpha \right\}$ is also closed. Is this correct?

For the second part I'm struggling more. Assume now that $H$ is closed, the set $H^c$ is open specifically I can write $$ \begin{array}{l} H^c = H_{+,\alpha} \cup H_{-,\alpha} \\ H_{+,\alpha} = \left\{x \in E : f(x) > \alpha \right\} = f^{-1}\left( \left(\alpha,+\infty\right) \right) \\ H_{-,\alpha} = \left\{x \in E : f(x) < \alpha \right\} = f^{-1}\left( \left(-\infty,\alpha\right) \right) \end{array} $$ Let's consider $H_{+,\alpha}$ (which is open) because it is an open set in a normed vector space, so for each $x_0 \in H_{+,\alpha}$ there's a ball $B_{r,x_0} = \left\{ x \in E : d_E(x,x_0) < r \right\} \subset H_{+,\alpha}$ (I'm exploiting here a topological fact more than what Brezis does, it does prove that the ball $B_{r,x_0}$ is in $H_{+,\alpha}$ by contradiction). Each $y \in B_{r,x_0}$ can be written as

$$ y = x_0 + rz\;\;, z \in B_{1,0} $$ From which we have $$ f(x_0 + rz) < \alpha \;\; \forall z \in B_{1,0} $$

This very last inequality make Brezis say that

Consequently $f$ is continuous...

Which I don't really understand why, from other reference I know that a linear functional is continuous if and only if it is continuous in $0$, but it doesn't seem to me that is mentioned in the book I've been talking about. My questions are

  1. Is my proof of "$f$ continuous $\Rightarrow H$ is closed" correct?
  2. How can be concluded that $f$ is continuous?
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    Brezis doesn't use the topological fact you mentioned because it is not true. – 2017-02-08
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    Why it's not true? – 2017-02-08
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    The burden of proof lies on you. Why is it true? – 2017-02-08
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    Because if $H_{+,\alpha}$ is an open set in a metric space $E$ then for each $x \in H_{+,\alpha}$ there's a basis $B$ element such that $x \in B \subset H_{+\alpha}$ (munkres book, how to define a topology using a base). The basis element in a normed vector space (which is a metric space) are open balls. That's the argument I would have used, why is this argument wrong? – 2017-02-08
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    Yes but why is $H_{+,\alpha}$ open? – 2017-02-08
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    Lol, you got me. I'll stick with Brezis. I've tried to expound the way, I'm ending up with it's same proof. – 2017-02-08

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$1.$ Yes; the preimage of a closed set by a continuous function is closed.

$2.$ Note that $f(x_0 + rz) = f(x_0) + rf(z)$, so $f(z) < \frac1r (\alpha - f(x_0)) :=M$ for all $z \in B[0,1]$ (it's $B(0,1)$ but an easy modification gives the closed ball). Hence $\sup_{z \in B[0,1]} |f(z)| \le M$, which means that $\|f\| \le M < \infty$ (in particular, $f$ is continuous).

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    I still don't understand, shouldn't be proved that $\forall \epsilon > 0$ there's a $\delta > 0$ such that $d(x,x_0) \delta \Rightarrow d(f(x),f(x_0)) < \epsilon$? – 2017-02-08
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    I got the bound, I don't get the final implication. – 2017-02-08
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    @user8469759 a linear map is continuous iff its operator norm is bounded. – 2017-02-08
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    I missed that point... Now it's ok. – 2017-02-08
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    However how the fact that $f(z) < \frac{1}{r}(\alpha - f(x_0))$ implies that $\sup_{z \in B[0,1]} |f(z)| < M$ couldn't happen that $f(z)$ is negative and small enough that $\sup_{z \in B[0,1]} |f(z)| > M$? – 2017-02-08
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    @user8469759 sorry, I got a little bit busy. I'll come back to this later. – 2017-02-08