Function Analysis with $f''(x) > 0$

The condition $f''(x) > 0$ for all $x \in \mathbb{R}$ means the function $f$ is strictly convex on $\mathbb{R}$. This also implies that the first derivative, $f'(x)$, is strictly increasing.

Statement Analysis

• 1. $f$ has a local minima
  This statement is NOT always correct. A strictly convex function may have a local minimum if $f'(x) = 0$ for some $x$. However, consider the function $f(x) = e^x$. Here, $f''(x) = e^x > 0$ for all $x$, but $f'(x) = e^x$ is never zero. Therefore, $f(x) = e^x$ has no local minimum.
• 2. There does not exist $x$ and $y, x \ne y$, such that $f'(x) = f'(y) = 0$
  This statement IS always correct. Assume for contradiction that there exist distinct $x$ and $y$ such that $f'(x) = f'(y) = 0$. Since $f$ is twice-differentiable, $f'$ is differentiable. By Rolle's Theorem applied to $f'$ on the interval $[x, y]$ (or $[y, x]$), there must exist a point $c$ between $x$ and $y$ such that $f''(c) = 0$. This contradicts the given condition that $f''(x) > 0$ for all $x$. Therefore, $f'(x)$ can be equal to 0 for at most one value of $x$.
• 3. $f$ has at most one global minimum
  This statement IS always correct. A strictly convex function ($f''(x) > 0$) can have at most one global minimum. If there were two distinct global minima, say at $x_1$ and $x_2$ with $f(x_1) = f(x_2) = m$, the line segment connecting $(x_1, m)$ and $(x_2, m)$ would lie strictly above the graph of $f$ between $x_1$ and $x_2$, contradicting strict convexity.
• 4. $f$ has at most one local minimum
  This statement IS always correct. A necessary condition for a local minimum at a point $x_0$ is $f'(x_0) = 0$. Since $f'(x)$ is strictly increasing (because $f''(x) > 0$), the equation $f'(x) = 0$ can have at most one solution. Thus, there can be at most one point where a local minimum occurs.

Conclusion

Based on the analysis, statements B, C, and D are always correct.