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From Brezis Functional Analysis pg.5

Proposition: Let $f$ be a linear functional on $E$ which does not vanish identically, $\alpha \in \mathbf{R}$. Then the affine hyperplane $H:=\{x\in E\,|\,f(x)=\alpha\}$ is closed if and only if $f$ is continuous.

My Proof: If $f$ is continuous and $\{h_k\}$ a sequence in $H$ then $\alpha = f(h_k)\rightarrow f(h) = \alpha$ as $k\rightarrow \infty$ and $H$ is closed.

Conversely let $H$ be closed and let $\{h_k\}$ be a sequence converging to $h\in H$. Then $f(h)=\alpha$ and $$ \lim_{k\rightarrow \infty}f(h_k) = \lim_{k\rightarrow \infty}\alpha = \alpha = f(h) $$ so $f$ is continuous. $\square$

Brezis does something far different for the second part and I was wondering if I have made a very fundamental mistake or he has chosen his method for pedagogical reasons. Thanks! Also any thoughts on my presentation and proof writing are appreciated.

Resolution: Guido has found the error. I have only proved that $f$ is continuous on $H$ where I actually need to prove that $f$ is continuous on all of $E$.

qualcuno
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krc
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    Your second part is not proving anything. The first equation therein is not justified and is a would-be consequence of the continuity of $f$, which is what is wanted to be proven. –  Feb 14 '20 at 04:15
  • You have shown that $f$ is continuous in $H$, but that will happen regardless of $H$ being closed simply because $f|_H$ is constant. What you have to prove is that $H$ being closed implies continuity of $f$ in its domain (and not only restricting it to $H$). – qualcuno Feb 14 '20 at 04:18
  • @Guido No they haven't. They have not proven continuity anywhere. –  Feb 14 '20 at 04:19
  • There is nothing to prove though, since $f|_H$ is constant, the function $f$ will be continuous, but they are (provided that ${h_k}_k \subset H$ which I think is implicit). They show that $f(h_k) \to f(h)$ when $h_k \to h$ (in $H$), but this is just because $f(h_k) = \alpha = f(h)$ if $h_k,h \in H$. – qualcuno Feb 14 '20 at 04:20
  • Ahh I see, Guido is correct that I have only shown continuous on $H$. @tora I think that the first equation is justified since $h\in H$ and $f(h)=\alpha$ for all $h\in H$ by the definition of the hyperplane. Brezis uses this fact in his proof as well. I think you're caught up on the fact that what I proved, while true, is trivial and not what I was trying to prove. – krc Feb 14 '20 at 04:32

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You have shown that $f$ is continuous in $H$, but that will happen regardless of $H$ being closed simply because $f|_H$ is constant. What you have to prove is that $H$ being closed implies continuity of $f$ in its domain (and not only restricting it to $H$).

Here's another proof:

Note that translation by a point is a homeomorphism, so $H - p$ will be closed for any $p \in E$. Now, if $f$ is zero the result follows, and otherwise by dimension $f$ is surjective and thus we can choose $x \in f^{-1}(\alpha)$. It follows that $\ker f = H - x$ is closed. Therefore, we can assume $\alpha = 0$ and $H = \ker f$.

Taking $y \in E \setminus \ker f$ , if $f$ were not bounded then we should have a sequence of unit vectors $x_n$ such that $f(x_n) \geq n$ for each $n \in \mathbb{N}$. But then it is

$$ f(y)f(x_n)^{-1}x_n -y \to y $$

with $f(y)f(x_n)^{-1}x_n -y \in \ker f$, and that contradicts the fact that $\ker f$ is closed.

qualcuno
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