I have come across a notion of orthogonality of two vectors in a normed space not necessarily inner product space. Two vectors $x$ and $y$ in a normed space are said to be orthogonal (represented $x\perp y$) if $||x||\leq ||x+\alpha y||,$ for every $\alpha,$ a scalar. 


1) What is the rational behind the definition above? I guess, it has got something to do with minimum overlap between $x$ and $y$. 


2) Is this unique generalization of the concept of orthogonality from inner product spaces?


Thank you. 



Orthogonality in non-inner product spaces
Concerning question 1: The rational is that in an inner product space $$x\perp y \Leftrightarrow \forall \alpha \in K: ||x||\leq ||x+\alpha y|| \qquad(K = \mathbb{R} \text{ or } K = \mathbb{C})$$
Now, if no inner product is available (but a norm), the idea is, to just take the right hand side as definition of orthogonality (call it $\perp\_1$). 


Concerning question 2: No, there are other -non-equivalent - generalizations as well. As an example, note that in an inner product space over the reals 
$$\langle x,y \rangle = \frac{1}{4}( ||x+y||^2 - ||x-y||^2).$$
Hence $x\perp y \Leftrightarrow ||x+y|| = ||x-y||$. So the definition 
$$ x\perp\_{\scriptstyle 2}\; y : \Leftrightarrow ||x+y|| = ||x-y||$$
generalizes the orthogonality from an inner product space to any normed space (over the reals).


Now let's show that $\perp\_1, \perp\_2$ aren't equivalent. Let $E = \mathbb{R}^2$ with norm $||(a,b)|| = \max(|a|, |b|)$. Then 


$\qquad (0,1) \perp\_2 (2,1)$ but **not** $(0,1) \perp\_1 (2,1)\quad$ (take $t=-1/4$)


$\qquad (1,1) \perp\_1 (2,0)$ but **not** $(1,1) \perp\_2 (2,0).$ 


