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I'm trying to prove that if $x$ and $y$ are in $\mathbb R^n$ such that $\langle x,e\rangle=0 $ where $e=(1,1,...,1)$ then

$$|\langle x,y\rangle |\le||x||_1\frac{y_{\max}-y_{\min}}{2}\qquad \text{for all} \quad y$$

I think that is necessary to use norms inequalities and Cauchy–Schwarz inequality or Hölder's inequality.

Can you help me, please?

J.G.
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Cristian Camilo Espitia Morill
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1 Answers1

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Proof is by induction. Suppose this holds for $n<N$. Since a rearrangement of $x_{k}^{\prime }s$ does not change the inequality we may suppose there exists $m$ such that $\ x_{1},x_{2},...,x_{m}$ are positive and $% x_{m+1},x_{m+1},...x_n$ are negative. (zero terms can be dropped). Now let $u_{k}=x_{k},v_{k}=y_{k}$ for $1\leq k\leq m$ and $u_{m+1}=-\sum _{k=m+1}^{n}x_{k}$. Let $v_{m+1}=\frac{\sum% _{k=m+1}^{n}x_{k}y_{k}}{\sum_{k=m+1}^{n}x_{k}}$. We apply induction hypothesis to the vectors $(u_{1},u_{2},...,u_{m+1})$ and $% (v_{1},v_{2},...,v_{m+1})$. Assume that at least two of the $x_{k}^{\prime }s$ are negative so that $m+1<n$. We can conclude, by induction hypothesis, that $$\left\vert \sum_{k=1}^{m+1}u_{k}v_{k}\right\vert \leq (\sum_{k=1}^{m+1}\left\vert u_{k}\right\vert )\frac{v_{\max }-v_{\min }}{2}$$. Note that $$v_{m+1}-v_{j}=\frac{\sum% _{k=m+1}^{n}x_{k}(y_{k}-v_{j})}{\sum_{k=m+1}^{n}x_{k}}=\frac{% \sum_{k=m+1}^{n}x_{k}(y_{k}-j_{j})}{\sum_{k=m+1}^{n}x_{k}}=% \frac{\sum_{k=m+1}^{n}\left\vert x_{k}\right\vert (y_{k}-y_{j})}{% \sum_{k=m+1}^{n}\left\vert x_{k}\right\vert }$$ $\leq y_{\max }-y_{\min } $. It follows from this that $v_{\max }-v_{\min }\leq y_{\max }-y_{\min }$ so we have $$\left\vert \sum_{k=1}^{m+1}u_{k}v_{k}\right\vert \leq (\sum_{k=1}^{m+1}\left\vert u_{k}\right\vert )\frac{y_{\max }-y_{\min }}{2}$$. Also note that $$\sum_{k=1}^{m+1}\left\vert u_{k}\right\vert =\sum_{k=1}^{m+1}x_{k}-\sum_{k=m+1}^{n}x_{k}=\sum% _{k=1}^{n}\left\vert x_{k}\right\vert $$ and $$\sum% _{k=1}^{m+1}u_{k}v_{k}=\sum_{k=1}^{m}u_{k}v_{k}-\sum% _{k=m+1}^{n}x_{k}\frac{\sum_{k=m+1}^{n}x_{k}y_{k}}{% \sum_{k=m+1}^{n}x_{k}}=\sum_{k=1}^{n}x_{k}y_{k}$$. This finishes the proof in this case. If $x_{1},x_{2},...,x_{n-1}$ are positive and $x_{n}$ is negative then $$\sum_{k=1}^{n}x_{k}y_{k}=\sum _{k=1}^{n-1}x_{k}y_{k}-(\sum_{k=1}^{n-1}x_{k})y_{n}=\sum _{k=1}^{n-1}x_{k}(y_{k}-y_{n})$$ so $$\left\vert \sum_{k=1}^{n}x_{k}y_{k}\right\vert \leq \sum_{k=1}^{n-1}\left\vert x_{k}\right\vert (y_{\max }-y_{\min })$$ and $$\sum_{k=1}^{n-1}\left\vert x_{k}\right\vert =\frac{1}{2}% \sum_{k=1}^{n}\left\vert x_{k}\right\vert $$. Hence the desired inequality holds in this case also. The inequality is obvious for $n=1$ so we have finished the proof.

Kavi Rama Murthy
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  • In this case, $\Vert x\Vert_1\geq 2|x_1|$. – W. mu Aug 15 '18 at 02:09
  • @W.mu I have edited the answer providing explicit values for $x_2,...,x_n$. – Kavi Rama Murthy Aug 15 '18 at 06:11
  • The definition of the norm $\Vert \Vert_1$ is $\Vert x\Vert_1=|x_1|+|x_2|$, not the Euclidean norm. – W. mu Aug 15 '18 at 07:57
  • @W.mu Where is the norm defined? I don't see it in the title or the question. If no norm on an Euclidean space is specified it is always taken as the usual norm. Normally the norm used is the one that comes from the inner product and since the inner product here is the usual one, the norm has to be the usual one. Unless the question is modified my answer is perfectly valid. – Kavi Rama Murthy Aug 15 '18 at 08:04
  • This is a standard notation $\Vert \Vert_p$. https://en.wikipedia.org/wiki/Lp_space#The_p-norm_in_finite_dimensions – W. mu Aug 15 '18 at 11:40
  • @W.mu After that mad mistake of not reading the right side properly I have come with a proof. Please take a look. – Kavi Rama Murthy Aug 20 '18 at 08:22