sharpe ratio, convert into convex function, not understand that constraint, [duplicate]

Question

I am reading about tranforming sharpe ratio into convex problem

After some following, its converted into min xTxy s.t. (u-rf e)x = 1

X=x+/k+

X+ = x/(u-rf e)x
K+ = 1/ (u-rf e)T x

In the book, it mention adding the normalizing constraint (u-rf)x = 1 does not affect the equation

I understand that f(x) is same as f(kx)

However, i get some questions here.

Why (u-rf)x = 1 can confirm its maximum sharpe ratio ?

So why this constraint

Answer 1

After reading about A Signal Processing Perspective on Financial Engineering.

I am trying to answer my question. But I am not sure that if I am correct or not. and show my question again.

I would like to ask why we add this (µ − rf1) = 1 constraint.

min xT Σx
subject to xT
(µ − rf1) = 1,
xT 1 > 0.

First, we do homogenizing x = kx, as f(x) = f(kx), so the objective function becomes wT Σw

x = x+ / k+
x+ = x / (µ − rf1)
k+ = 1 / (µ − rf1)

this paramater given in max, so in min k become (µ − rf1)

after x = x+ / k+, it will be same as x

as x = k+x+

as we scale constraint wT 1 = 1 can be relaxed to wT 1 > 0 so we can assume one arbitrarily set as wT(µ − rf1) = 1

k = (µ − rf1) `in min that` wT(µ − rf1) = 1