MHB Is the Minimum of a Strictly Convex Function Unique at Its Critical Point?

[JS10]

Hello,
Can you please help me to solve this exercise:

Let f a function that satisfies:
- f is class C2 and strictly convex (f'' (x) > 0).
- There is x*, f' (x*) = 0.
Question is: prove that the minimum of f is reached in x* and it's unique? Using the descent gradient method (build a sequence (Xn)/ Xn+1 = Xn - γ f′(Xn)).

Thanks.

 

[Jhoneine]

Answer:To prove that the minimum of f is reached at x* and it is unique, we will use the descent gradient method. We will construct a sequence (Xn) where Xn+1 = Xn − γ f′(Xn). Since f is strictly convex and has a unique minimum at x*, we know that the function is monotonically increasing and it reaches its minimum at x*. Thus, if we start at any point X0 and use the descent gradient method to descend along the gradient of the function, we will eventually reach the point x* where f'(x*) = 0.We can see this by noting that the descent gradient method updates the current point Xn according to the equation Xn+1 = Xn − γ f′(Xn). Since f is strictly convex, we know that f''(x) > 0 for all x in the domain of f. This means that the derivative f′(x) is always decreasing, so the descent gradient method will always move in the direction of the negative gradient, which is the direction of steepest descent. As we keep descending along the negative gradient, we will eventually reach the point x* where f'(x*) = 0 and the function reaches its minimum. Therefore, the minimum of f is reached at x* and it is unique.

 

[soloenergy]

Sure, I'll try my best to help you with this exercise. First, let's define the descent gradient method, also known as gradient descent. This method is used to find the minimum of a function by iteratively updating a guess for the minimum using the gradient (or slope) of the function at that point.

Now, let's look at the problem at hand. We have a strictly convex function f that is also C2, meaning it is twice continuously differentiable. We also know that there is a point x* where the derivative of f is 0, or f'(x*) = 0. Our goal is to prove that the minimum of f is reached at x* and that it is unique.

To do this, we will use the descent gradient method. We will start with an initial guess x0 and then follow the sequence Xn+1 = Xn - γ f'(Xn), where γ is a small positive number called the learning rate. This sequence will iteratively update our guess for the minimum until we reach a point where the derivative is 0, indicating that we have reached the minimum.

Now, let's assume that there exists another point x' where the minimum of f is also reached. Since f is strictly convex, we know that the derivative of f is strictly increasing. This means that as we move away from x*, the derivative of f will become more positive. Similarly, as we move towards x*, the derivative of f will become more negative.

Since x' is another point where the minimum of f is reached, we know that f'(x') = 0. This means that the derivative of f at x' is 0, but since x' is not x*, we know that f'(x') ≠ 0. This contradicts the fact that f'(x*) = 0. Therefore, our assumption that there exists another point x' where the minimum of f is reached is false, and we have proven that the minimum of f is unique and is reached at x*.

To conclude, we have proven that the minimum of f is reached at x* and that it is unique, using the descent gradient method. This method can be applied to any strictly convex function to find its minimum. I hope this explanation was helpful. Let me know if you have any further questions. Good luck with your exercise!