Simple question.
The function g of a single variable is defined by g(x) = f(ax + b), where f is a concave function of a single variable that is not necessarily differentiable, and a and b are constants with a ≠ 0. (These constants may be positive or negative.) Either show that the function g is concave, or show that it is not necessarily concave. [Your argument must apply to the case in which f is is not necessarily differentiable.]
I know that the condition for concavity is:
g((1-α)x1 + αx2) ≥ (1-α)g(x1) + αg(x2) for all x1 and x2 where x1 < x2, and α ∈ [0,1]
Because g(x) = f(ax + b) g((1-α)x1 + αx2) = f(a((1-α)x1 + αx2) + b) g((1-α)x1 + αx2) = f((1-α)(ax1+b) + α(ax2+b)) g((1-α)x1 + αx2) ≥ αf(ax2 + b) + (1−α)f(ax1 + b) (by the concavity of f ) = αg(x2) + (1 − α)g(x1).
Why is f((1-α)(ax1+b) + α(ax2+b)) ≥ αf(ax1 + b) + (1−α)f(ax2 + b) the condition for concavity of f(ax+b)?
I thought that at x = (1-α)x1 + αx2, the value of f(ax+b) = f(a((1-α)x1 + αx2) + b) and therefore this value must be greater than the corresponding value on the line connecting the endpoints x1 and x2, which is αf(x2) + (1−α)f(x1), or αf(ax2 + b) + (1−α)f(ax1 + b).
Why is it necessary to show that f((1-α)(ax1+b) + α(ax2+b))?
Thank you in advance for your help.
Sincerely, OMAX
$$g(αx_1 + (1−α)x_2) = f (a(αx_1 + (1−α)x_2) + b) = f (α(ax_1 + b) + (1−α)(ax_2 + b)) ≥ α f (ax_1 + b) + (1−α) f (ax_2 + b) (\text{by the concavity of} f ) = αg(x_1) + (1 − α)g(x_2)$$.