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E is a normed linear space . i have two questions Q1 why the complement of H is nonempty Q2 How then the functional is continuous?? Thanks

Roba
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  • $H^c$ is nonempty since it is assumed that $f\neq 0$ identically in the definition of a hyperplane. – Jas Ter Dec 02 '14 at 15:32
  • $f$ is continuous since its norm is estimated and finite. – Jas Ter Dec 02 '14 at 15:33
  • your first comment is what have been written in the proof but i cann't see it? – Roba Dec 02 '14 at 15:43
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    functional is bounded if there exist K>0 such that ||Tx||<=K||x|| – Roba Dec 02 '14 at 15:44
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    my interpretation is since unit ball of normed linear space is symmetric then f(-u) is also <(alpha-f($x_0$))/r then the functional is bounded on some neighbourhood of zero and i can prove it is cts at zero then it is continuous at all points of the space,is that right? – Roba Dec 02 '14 at 15:50
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    If $f$ does not vanish identically, you can use Hahn-Banach and show the existence of some $x$ such that $f(x)\neq \alpha$. – Jas Ter Dec 02 '14 at 17:39
  • Yes, continuity at zero is equiv to continuity for linear functionals, which again is equiv to boundedness. – Jas Ter Dec 02 '14 at 17:39
  • do you mean that my interpretation is true for the part of continuity ?? How can i use hahn-banach to show that f(x) not equal alpha for some x – Roba Dec 02 '14 at 18:26
  • please,help me specially in the point of how to use hahn-banach to show that f(x) not equal alpha for some x. Thanks – Roba Dec 02 '14 at 18:33
  • My comment regarding Hahn-Banach is irrelevant/not correct. See answer below. – Jas Ter Dec 02 '14 at 19:51
  • i don't get the "f is continuous" part – 0212user Oct 25 '17 at 22:48

1 Answers1

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$Q1$: There is an implicit assumption in the theorem, and that is that $f$ is not identically zero. If $f=0$ then $\{x|f(x)=\alpha\}$ is either empty or the whole Banach space $X$, neither of which is a hyperplane.

$H^c =\{ x|f(x)\neq \alpha\}$. Suppose $H^c$ is empty. Then $f(x)\neq \alpha$ is impossible, so $f(x) = \alpha$ for all $x$. This contradicts linearity of $f$. (Easy to check.)

$Q2$: For linear functionals $f$ over a Banach space $X$, continuity at $x\in X$ is equivalent to contiuity at $0\in X$, which is equivalent to local boundedness at $0\in X$, i.e., that $$ \|f\| = \sup_{z\in B(0,1)} \frac{|f(z)|}{\|z\|} < +\infty. $$

Given that $ f(x_0 + r z) < \alpha$ for all $z \in B(0,1)$ we have also that $f(x_0 - r z)<\alpha$, since $-z\in B(0,1)$ if and only if $z\in B(0,1)$. Thus, $-f(z) = f(-z) < \tfrac{1}{r}(\alpha - f(x_0))$, as well as $f(z) < \tfrac{1}{r}(\alpha - f(x_0))$, and $$ |f(z)| < \frac{1}{r}(\alpha - f(x_0)) $$ for all $z\in B(0,1)$. It follows that $$ \|f\| \leq \frac{1}{r}(\alpha - f(x_0)). $$

Jas Ter
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