Problem about Hahn-Banach Theorem

Question

Could someone help me with the following problem?

Let be $X$ a real normed space , $f,g\in{X^*}$ ($\|f\|=\|g\|=1$) and $0<r\leq1$ such that $|f(x)|\leq{r}$ $\forall{x\in{\ker(g)}},\|x\|\leq{1}$. Prove that either $\|f-g\|\leq{2r}$ or $\|f+g\|\leq{2r}$.

I am trying to use Hahn-Banach Theorem to $f_{\ker(g)}$, but I don't get nothing .

Thanks.

How can $f$ be bounded when it's linear unless $f \equiv 0$? — md2perpe, Jan 04 '18 at 18:08
@md2perpe I take it that when you say "$f$ bounded" you mean $|f(x)|\le c$ for all $x$. Why do you think that's so? — David C. Ullrich, Jan 04 '18 at 18:17
I saw $|f(x)|\leq{r}$ in a subspace, but I missed the restriction to $|x| \leq 1.$ Sorry for that. — md2perpe, Jan 04 '18 at 19:56
Are you sure that this is true? Check $X=\mathbb R^2$, $g(x,y)=y$ and $f(x,y)=(x+y)/\sqrt{2}$. — Jochen, Jan 05 '18 at 11:13
Yes, @Jochen you're right, so what can we say if we change $r$ for $2r$?.(I've just edited again the question) — mathlife, Jan 05 '18 at 11:24
Did you try $X=\mathbb R^2$? Perhaps with different norms $|(x,y)|_t^2=x^2+ty^2$ for a parameter $t>0$. — Jochen, Jan 05 '18 at 13:06
Yes, @Jochen I have been looking for a counterexample but it sems to me that this is true — mathlife, Jan 05 '18 at 13:41

score 2 · Answer 1 · edited Jan 09 '18 at 12:52

2

Well, as long as we're willing to change the question, it's not hard to show that $\|f+g\|\le 1+r$ or $\|f-g\|\le 1+r$.

Let $Z$ be the kernel of $g$. We have $\|f|_Z\|\le r$, so Hahn-Banach shows that there exists $F$ with $F|_Z=f|_Z$ and $\|F\|\le r$.

Now $F-f$ vanishes on the kernel of $g$, so there exists $c$ with $$F-f=cg.$$Note that $|c|=\|F-f\|\le 1+r$.

If $0\le c\le 1+r$ then $$\|f+g\|\le\|f+cg\|+|c-1|\le r+1.$$ Similarly if $-1-r\le c\le 0$ then $\|f-g\|\le 1+r$.

edited Jan 09 '18 at 12:52

Martin Sleziak

53,687

answered Jan 06 '18 at 01:27

David C. Ullrich

89,985

Why does $\operatorname{Ker} g \subseteq \operatorname{Ker} (F - f)$ imply that $F - f = cg$, for some scalar $c$? – mechanodroid Jan 06 '18 at 09:21
1

@mechanodroid If $g=0$ then $F-f=0$. If $g\ne0$ choose $x_0$ with $g(x_0)=1$ and let $c=(F-f)(x_0)$. (Note that $X$ is spanned by $Z,x_0$...) – David C. Ullrich Jan 06 '18 at 13:57

Problem about Hahn-Banach Theorem

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