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Let $v_0=(v_{0x},0),v_1=(v1x,0),w_0=r_0e^{i\theta_0},w_1=r_1e^{i\theta_1}\in \mathbb{R}^2$ be any four non-zero vectors such that:

  • $v_{0x}>0$ and $v_{1x}>0$
  • $0<\theta_0<\pi$ and $0<\theta_1<\pi$
  • the following inequalities are satisfied: $$\displaystyle{\frac{||v_1||}{||v_0||}> \frac{||w_1||}{||w_0||}}\quad \textit{and }\frac{||v_1||}{||v_0||}> \frac{||v_1+w_1||}{||v_0+w_0||}$$

Now define, for every $t\in [0,1]$, the following vectors:

$$\displaystyle{v_t=((1-t)v_{0x}+tv_{1x},0)}$$

$$\displaystyle{w_t=(r_0(1-t)+tr_1)e^{i((1-t)\theta_0+t\theta_1)}}$$

Is the following inequality

$$\displaystyle{\frac{||v_{t_1}||}{||v_{t_0}||}\ge \frac{||v_{t_1}+w_{t_1}||}{||v_{t_0}+w_{t_0}||}}$$

verified for every $t_0,t_1\in [0,1],t_0<t_1$?

If the answer is no, I'm searching for additional conditions on $v_1,v_2,w_1,w_2$ which will guarantee that this inequality is always verified.


My attempt: I've tried many examples without findind a counterexample. But I couldn't find a general proof.

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  • $\begingroup$ What about the case $v_1=-w_1$ and $m=n=1$? $\endgroup$ – Paolo Leonetti Jun 21 '17 at 15:51
  • $\begingroup$ @PaoloLeonetti You're right, there is a problem, I'll fix it $\endgroup$ – User29983 Jun 21 '17 at 15:54
  • $\begingroup$ Are you sure you don't mean to be asking about convex combinations, e.g., $v_t = (1-t)v_1 + t v_2$? I can't imagine why a deformation would be a linear combination in both magnitude and angle. $\endgroup$ – adfriedman Jun 21 '17 at 21:02
  • $\begingroup$ @adfriedman yes, I'm sure $\endgroup$ – User29983 Jun 21 '17 at 22:34
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The statement (of the original question) is not true. Take $v_1=e^{2\pi i/3}=(-1/2,\sqrt{3}/2)$, $v_2=e^{4\pi i/3}=(-1/2,-\sqrt{3}/2)$, $w_0=w_1=(1,0)=e^0$.

Then

  • $v_1$ and $w_1$ are linearly independent
  • $v_2$ and $w_2$ are linearly independent
  • the following inequalities are satisfied: $$\displaystyle{\frac{||v_2||}{||v_1||}=1\ge 1= \frac{||w_2||}{||w_1||}}\quad \text{and}\quad \frac{||v_2||}{||v_1||}=1\ge 1=\frac{||v_2+w_2||}{||v_1+w_1||}$$

For all $t$, we have $w_t=(1,0)$. For $t=1/4$ we have $v_t=e^{i((1-t)2\pi/3+t4\pi/3)}=e^{5\pi/6}=(-\sqrt{3}/2,1/2)$. Hence $v_t+w_t=(1-\sqrt{3}/2,1/2)$ and so $\|v_t+w_t\|=\sqrt{2-\sqrt{3}}<1$. Set $t_1=1$ and $t_0=1/4$. Then $$ \frac{||v_{t_1}||}{||v_{t_0}||}=1<\frac{1}{\sqrt{2-\sqrt{3}}}=\frac{||v_{t_1}+w_{t_1}||}{||v_{t_0}+w_{t_0}||} $$

I don't see any conditions you can put on $v_1,v_2,w_1,w_2$ such that the inequality is satisfied. What I did was to find the vectors such that you have equality for $t=0$ and for $t=1$. Then there must be an intermediate value where your inequality fails.

${\bf EDIT:}$ As for the second version of the question we also find a counterexample, that is, $v_0(v_{0x},0),v_1=(v_{1x},0),w_0,w_1$ such that

  • $v_{0x}>0$ and $v_{1x}>0$

  • $0<\theta_0<\pi$ and $0<\theta_1<\pi$

  • the following inequalities are satisfied:

$$\displaystyle{\frac{||v_1||}{||v_0||}> \frac{||w_1||}{||w_0||}}\quad \textit{and }\frac{||v_1||}{||v_0||}> \frac{||v_1+w_1||}{||v_0+w_0||}$$

and such that for $t_1=1/2$ and $t_0=0$ we have

$$\displaystyle{\frac{||v_{t_1}||}{||v_{t_0}||}< \frac{||v_{t_1}+w_{t_1}||}{||v_{t_0}+w_{t_0}||}}$$ where

$$\displaystyle{v_t=((1-t)v_{0x}+tv_{1x},0)}$$

$$\displaystyle{w_t=(r_0(1-t)+tr_1)e^{i((1-t)\theta_0+t\theta_1)}}$$ with $\theta_1=\pi/2$ and $\theta_0=\pi/2+\arctan(5/\sqrt{299})$.

In fact, set $v_{0x}=4$, $v_{1x}=3.04$ $w_1=(0,2)$ and $w_0=\frac{2}{9}(-5,\sqrt{299})$.

Then $\frac{||v_1||}{||v_0||}=\frac 34+0.01=0.76$, $\frac{||w_1||}{||w_0||}=\frac 34$, and $\frac{||v_1+w_1||}{||v_0+w_0||}=\frac{3\sqrt{13.2416}}{4\sqrt{13}}=0.756937...<0.76$.

But for $t_1=1/2$ and $t_0=0$ we have $||v_{t_1}||=3.52$, $||v_{t_0}||=4$, $||v_{t_0}+w_{t_0}||=\frac 43\sqrt{13}$, $||v_{t_1}+w_{t_1}||=\frac{1}{2} \sqrt{85-84 \sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{5}{\sqrt{299}}\right)\right)}=4.27834...$

and so

$\frac{||v_{t_1}||}{||v_{t_0}||}=0.88$ and $\frac{||v_1+w_1||}{||v_0+w_0||}=0.889949...$, so the inequality is not satisfied.

The strategy for finding a counterexample was to find $v_0,v_1,w_0,w_1$ such that
$$(1)\quad\quad\quad\frac{||v_1||}{||v_0||}= \frac{||w_1||}{||w_0||}\quad \text{and}\quad\frac{||v_1||}{||v_0||}= \frac{||v_1+w_1||}{||v_0+w_0||},$$ then for all $t\in[0,1]$ we should have the inequalites $$ \frac{||w_1+v_1||}{||v_1||}\le \frac{||w_t+v_t||}{||v_t||}\le \frac{||w_0+v_0||}{||v_0||}=\frac{||w_1+v_1||}{||v_1||}, $$ but we find some $t$ that doesn't satisfies the inequalites, and then disturb the initial data a bit, so that we have inequalities in (1) but the same $t$ still contradicts the desired conclusion.

This strategy takes advantage of the fact that the deformation is not very well behaved.

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  • $\begingroup$ Thank you. If I understand correctly your counterexample works only if the hypothesis inequalities are equalities. Then one could put the additional condition that the inequalities are strict: $\frac{||v_2||}{||v_1||}> \frac{||w_2||}{||w_1||}$ and $\frac{||v_2||}{||v_1||}> \frac{||v_2+w_2||}{||v_1+w_1||}$. Would it still be not true? $\endgroup$ – User29983 Jun 24 '17 at 8:43
  • $\begingroup$ No, you can sum small perturbations to the vectors in order to get strict inequalities, and the conclusion is still wrong. Taking the inequalities as equalities is just the easiest way to obtain counterexamples. $\endgroup$ – san Jun 24 '17 at 17:56
  • $\begingroup$ Thank you. I don't understand how these small perturbation should be applied in order to obtain a counter example for the strict inequalities. Consider the case $\frac{||v_2||}{||v_1||}>\frac{||w_2||}{||w_1||}>1$ and $\frac{||v_2||}{||v_1||}>\frac{||w_2+v_2||}{||w_1+v_1||}>1$. Can you please write explicitly how can one obtain a counterexample? $\endgroup$ – User29983 Jun 24 '17 at 18:42
  • $\begingroup$ For $\varepsilon$ small enough (for example $\varepsilon=1/100$) define $\bar v_2=v_2(1+2\varepsilon)$, $\bar w_2=w_2(1+\varepsilon)$. Then the inequalities are strict, and you get still a contradiction for $t=1/4$. $\endgroup$ – san Jun 25 '17 at 0:10
  • $\begingroup$ Thank you, it's clearer now. But if I understand correctly then does your counterexample work only for $\frac{||v_2||}{||v_1||}$ and $\frac{||w_2||}{||w_1||}$ close to 1? Can't we impose $\frac{||v_2||}{||v_1||}$ and $\frac{||w_2||}{||w_1||}$ to be $>>1$ to force my claim to work? $\endgroup$ – User29983 Jun 25 '17 at 11:22

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