Can a uni-modal function have a sub range of horizontal y-values? I'm now trying to figure out how the golden section search method works, one of the conditions for using this method is that the function you work on must be "uni-modal" and we work in a defined range (let the range of 'x' be [a0,b0]), so that if you have two x-values on the range you can get their corresponding y-values and start a comparison; "let the first point is x1 and the second is x2 where x1 < x2, then if y1 < y2 , the minimum value of the function f(x) must be included inside the range [a0,x2] .... and if y1 >= y2 then the minimum value of the function must be included inside the range [x1,b0]" , now you can see the "=" sign in (y1 >= y2).
so, my question here is "does a unimodal function can have a sub range inside its total range where all the values inside the sub range are equal (a range of horizontal straight line f(x) values), because if that is right then the condition stated here (y1 >= y2) is not true.
this is an image that can make my words more comprehensible.

thanks in advance
 A: You are trying to find $\min\limits_x f(x)$, so you are looking for a single value $x_{\min}$ for which $\forall x:f(x)\ge f(x_{\min})$, using the fact that $f(x)$ is a weakly decreasing function below $x_{\min}$ and a weakly increasing function above $x_{\min}$ 
There may be several values of $y$ for which $f(y)= \min\limits_x f(x)$, in which case there could be more than one minimiser.  
You are correct that if $f(a_1) \ge f(b_1)$  then it may be possible that some of the minimisers could be less than $a_1$.  But that does not matter if what you are looking for is a particular minimiser $x_{\min}$ rather than all the minimisers $\{y: f(y)= \min\limits_x f(x)\}$, because in this situation and looking for a particular minimiser you would have $a_1$ and $b_1$ as minimisers too and so some of the minimisers must be greater than or equal to $a_1$.  Combine this with every other case of  $f(a_1) \ge f(b_1)$, and the original statement is correct with a minor adjustment to the article:  

If $f(a_1) \ge f(b_1)$, then a minimiser is located in the range $[a_1, b_0]$ 

