Let $E$ be a normed space and let a vector $u\in E$ and a vector $v\in E$ with $||v||=1$ that satisy $||u+v||\geq 2-\varepsilon $ for a $\varepsilon >0$ be given. Show that for all $\alpha ,\beta >0$ the following holds:

$$||\alpha u+\beta v||\geq (1-\varepsilon )(\alpha +\beta )$$.

I have been stuck with this question for days and this is how far I've come:

$||\alpha u+\beta v||\leq ||\alpha u||+||\beta v||\\ =|\alpha |||u||+|\beta |||v||\\ =\alpha ||u||+\beta ||v||\\ =\alpha ||u||+\beta \\ \Rightarrow ||\alpha u+\beta v||\leq \alpha ||u||+\beta$


$||u+v||\leq ||u||+||v||=||u||+1\\ \Rightarrow 2-\varepsilon\leq ||u+v||\leq ||u||+1\\ \Rightarrow 2-\varepsilon \leq ||u||+1\\ \Rightarrow 1-\varepsilon \leq ||u||\\ \Rightarrow (\alpha +\beta )(1-\varepsilon )\leq ||u||(\alpha +\beta )$

Which seems to be very close, but it's not yet there and I can't get any longer.

A combination of the above (assuming that the inequality I'm supposed to show holds true) gives:

$\alpha ||u||+\beta ||u||\leq \alpha ||u||+\beta \\ \beta ||u||\leq \beta \\ ||u||\leq 1$

The definition of a norm in combination with what's given in the question should also lead to $0<\varepsilon \leq 2$.

But that's it, I just can't prove it. I have managed to show that it holds true for $\varepsilon =2$, but the assumption should be for a given $\varepsilon >0$, not one that I assume myself.

Any ideas?

  • $\begingroup$ If $\varepsilon \geq 1$ what you should prove is trivially true by the definition of a norm, so you can assume $0 < \varepsilon < 1$. $\endgroup$ – M.U. Sep 21 '15 at 18:10
  • $\begingroup$ I don't really see how that helps, or why it would be trivial for that matter. The definition of a norm only says that $||\alpha u+\beta v||\geq 0$, but I can't just assume that it's greater than $(1-\varepsilon )(\alpha +\beta )$ just because $(1-\varepsilon )(\alpha +\beta )\geq 0$. Or did I misinterpret you? $\endgroup$ – curroergosum Sep 21 '15 at 18:24
  • $\begingroup$ If $\varepsilon \geq 1$ then the right hand side of your "to-prove-inequality" is zero or even negative. However I need to admit that this was just the first thing i noticed without thinking if it may help. $\endgroup$ – M.U. Sep 21 '15 at 18:28
  • $\begingroup$ Ah, my bad, I just thought of it as $(\varepsilon -1)$, but now when I got it right it might help, but I'll have to look into it :) $\endgroup$ – curroergosum Sep 21 '15 at 18:41
  • $\begingroup$ So, I tried to go through it and it's fine for the upper limit ($\varepsilon=1$), but not for the lower limit ($\varepsilon=0$)... $\endgroup$ – curroergosum Sep 21 '15 at 19:08

Let $E$ be the 2-d Euclidean space with the Euclidean norm, $v = (1,0)$ and $u = (0,\sqrt{3})$. Then $\| u + v\| = 2$. Fix $\epsilon = 0.000001$. Apparently $\|v\| = 1$ and $\| u + v\| \geqslant 2 - \epsilon$.

Now set $\alpha = \frac14$ and $\beta = \frac34$, then $\alpha + \beta = 1$ and

$$ \|\alpha u + \beta v\| = \|(\frac34, \frac{\sqrt{3}}{4})\| = \frac{\sqrt{3}}{2} < (1 - \epsilon)\cdot (\alpha + \beta).$$

  • $\begingroup$ But $||\alpha u+\beta v||$ should be GREATER than $(1-\varepsilon )(\alpha +\beta )$. And in addition, I'm not allowed to fix $\varepsilon $ and neither $\alpha $ and $\beta $, because it should hold true for all $\alpha $ and $\beta $. $\endgroup$ – curroergosum Sep 22 '15 at 4:55
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    $\begingroup$ Yes. It is a counterexample. Your proposition is false. $\endgroup$ – corindo Sep 22 '15 at 5:04
  • $\begingroup$ But you cannot prove that it's wrong with a counterexample with a fixed $\varepsilon$, because it doesnt say that the inequality holds true for ALL $\varepsilon>0$, just that it holds true for SOME $\varepsilon>0$. I know that the inequality is true, that's given (and I have double-checked it with my professor), I just don't know how to show it, since I'm stuck... $\endgroup$ – curroergosum Sep 22 '15 at 14:30
  • $\begingroup$ You and your professor can go on and on with it. Or you can show my answer to him. Maybe he knows some logic. $\endgroup$ – corindo Sep 23 '15 at 1:14
  • $\begingroup$ But if you in your example fix $\varepsilon=0.9999999$ instead, then the proposition holds true and that's my point - it only needs to hold true for one epsilon in order for the proposition to be true (but for any $\alpha, \beta$ and not for fixed vectors). I'm a post-grad student, if that says anything about the level. We are always supposed to prove and prove things wrong without fixing variables/constants. $\endgroup$ – curroergosum Sep 23 '15 at 6:50

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