Does it always happen $\|ax+by\|=1 \implies |a+b|\leq 1?$ for $\|x\|=\|y\|=1$ (where $a, b\in \mathbb{R}$)

Question

I am new in functional analysis. I was reading a theory related to unit ball in Banach space. While studying the norm attainment set of points I met with a confusion:

For a finite dimensional real Banach space $\mathbb{X}$ and for $\|x\|=\|y\|=1$, do always $\|ax+by\|=1 \implies |a+b|\leq 1? $ (where $a, b\in \mathbb{R}$)

I think the answer is yes. Also I have a doubt on whether the statement is true in any normed linear space. I have tried with my limited knowledge to prove it by contradiction i.e., by considering $|a+b|>1$ trying to find two distinct elements $x, y$ with $\|x\|=\|y\|=1$ such that $\|ax+by\|\ne 1$ but I could not able to proceed. Please help me. Thank you in advance.

Thank you all for the quick suggestions but all the examples are in Hilbert space. What will happen if $\mathbb{X}$ is not a Hilbert space? Is the statement still wrong?

No, take $x = (1,0)$, $y = (0,1)$ and $a = b = \frac{1}{\sqrt{2}}$, then $||ax+by|| = 1$ but $|a+b| = \frac{2}{\sqrt{2}} = \sqrt{2} > 1$ — StiftungWarentest, Feb 15 '23 at 15:35
And an even dumber counterexample in $\Bbb R$: $x=1$, $y=-1$, $a=2$, $b=1$. — TZakrevskiy, Feb 15 '23 at 15:39
Hi,wellcome to MSE : does it help$$1=(a^2+b^2)(x^2+y^2)\geq(ax+by)^2, $$ ? — Khosrotash, Feb 15 '23 at 15:41
Thank you all for your valuable comment on this. It really helps me a lot. — Tutun, Feb 15 '23 at 20:08

score 1 · Accepted Answer · answered Feb 17 '23 at 11:38

1

TZakrevskiy's comment shows you how to construct a counterexample in any normed space.

Let $y = -x$ with $\Vert x \Vert = 1$. Then $$\Vert ax+by \Vert = \Vert (a-b)x\Vert = |a-b|.$$ Thus, we can let $b = a - 1$ with $a>1$ arbitrary to get a counterexample.

answered Feb 17 '23 at 11:38

Ennar

23,082

Got it @Ennar. Thank you very much. – Tutun Feb 17 '23 at 12:24
@Tutun, you are welcome. Here you can see something broader, restricting any norm to one-dimensional real subspace gives you $\mathbb R$ isometrically. So, if you can find counterexample in $\mathbb R$, you can find it in any normed space. – Ennar Feb 17 '23 at 12:33
Wow what a nice observation @Ennar!! Can you suggest any books or something so that I can develop my thinking like the way you have developed?? It will be very helpful for me. Thank you in advance. Any suggestion is welcomed. – Tutun Feb 17 '23 at 12:39
@Tutun, I'm not a functional analyst so unfortunately I can't recommend you any books in English, but you can easily find book recommendations on this site. For example, there is this question: https://math.stackexchange.com/q/7512/122131. – Ennar Feb 17 '23 at 12:45
Thank you @Ennar. Thank you very much for your suggestion. – Tutun Feb 17 '23 at 12:48

Does it always happen $\|ax+by\|=1 \implies |a+b|\leq 1?$ for $\|x\|=\|y\|=1$ (where $a, b\in \mathbb{R}$)

1 Answers1