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Let $(X,\|.\|)$ be a Banach space over the real line. Let $x\neq 0$ and $y\neq 0$ be in $X$, then $x$ is said to be orthogonal to $y$ if $\|x+\lambda y\|\geq\|x\|$ for every real number $\lambda$. It is said that given any $x$ and $y$ there exists a real number $\alpha$ such that $x$ is orthogonal to $\alpha x + y$. Can anyone tell how?

For reference see Corollary 2.2 in this article. I am not able to prove the existence in the corollary, but am able to prove the rest of it.

I can see it diagrammatically, that is we need to show that $x$ is orthogonal to a point in the line passing through $y$ and in the direction of $x$. But I am not able to find this $\alpha$ rigourously

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    $\begingroup$ This is called Birkhoff, or Birkhoff-James orthogonality. A good survey on the topic, with a lot of references is "On Birkhoff orthogonality and isosceles orthogonality in normed linear spaces" by Javier Alonso, Horst Martini and Senlin Wu. The specific result that you cite (and some others) were obtained by James in 1947. $\endgroup$ – erz Jan 20 at 9:45
  • $\begingroup$ @erz Yes you are correct. I have made a few changes in my questions for clarity. $\endgroup$ – user534666 Jan 20 at 10:22
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Consider the plane generated by $x$ and $y$. Now draw a tangent line to the sphere $\{z:\|z\|=\|x\|\}$ at point $x$. It has the form $\{x+tw|t\in \mathbb{R} \} $, and $w$ may be chosen of the form $\alpha x+y$. $x$ is orthogonal to $w$. This all works if $x$ and $y$ are not collinear; if they are the statement is formally false.

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  • $\begingroup$ If they are collinear, isn't it formally true, by choosing $\alpha$ such that $\alpha x + y = 0$? $\endgroup$ – user44191 Jan 26 at 22:35
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    $\begingroup$ @user44191 According to OP, formally both orthogonal vectors must be non-zero. $\endgroup$ – Fedor Petrov Jan 27 at 0:29
  • $\begingroup$ Thank you; I missed the "nonzero" (expecting it in a different spot). $\endgroup$ – user44191 Jan 27 at 1:25

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