Question

Let f be strictly convex. If f has a minimum, show that it is unique. (Hint: assume there are two minima x1, x2 and derive a contradiction using the definition of convexity.)

Answer 1

Answer: Ok, i will use the hint provided.

We call x1 and x2 to te two mínima of f, that is $\frac{df}{dx} (x1) = 0, \frac{df}{dx} (x2) = 0$.

The convexity condition says that, if f is differentiable, then the graph of f(x) lays above all the tangents between X and Y, if Y>X then

$f(Y) \geq f(X) + f'(X)*(Y-X)$ where $f'(x) = \frac{df}{dx} (x)$.

then, if we took Y = x1 and X = x2, we have $f(x1) \geq f(x2)$ because f'(x2) = 0.

now if we took Y = x2 and X = x1,  we have $f(x2) \geq f(x1)$ because f'(x1) = 0.

Then, x1 = x2. Which implies that we only have one minima.