技术问询：满足特定条件的[0;∞[上的函数f(x)是否为凸函数？

Great question! The short answer is no—a function satisfying all four listed conditions does not have to be convex. Let's break this down with a concrete counterexample that meets every requirement but fails to be convex.

Counterexample Function #

Consider the function:

f(x) = 1.1x + e^{-x} + \frac{1}{2}\sin^2(x)

Defined for all (x \in [0, \infty)). Let's verify each condition is satisfied:

1. (f(x) \geq x) for all (x \geq 0)
   Rearranging gives (0.1x + e^{-x} + \frac{1}{2}\sin^2(x) \geq 0). Every term here is non-negative for (x \geq 0): (0.1x \geq 0), (e^{-x} > 0), and (\sin^2(x) \geq 0). So this condition holds.

2. Has a global minimum on ([0, \infty))
   The function is strictly increasing (we'll confirm this next), so its smallest value occurs at the left endpoint (x=0), where (f(0) = 0 + 1 + 0 = 1). This is the global minimum.

3. Third derivative exists on ([0, \infty))
   Let's compute the derivatives step by step:

   • First derivative: (f'(x) = 1.1 - e^{-x} + \sin(x)\cos(x))
   • Second derivative: (f''(x) = e^{-x} + \cos(2x))
   • Third derivative: (f'''(x) = -e^{-x} - 2\sin(2x))
     All three derivatives are defined and continuous for all (x \geq 0), so the third derivative exists everywhere on the interval.

4. Strictly increasing on ([0, \infty))
   For all (x \geq 0):

   • (1.1 - e^{-x} \geq 1.1 - 1 = 0.1) (since (e^{-x} \leq 1) for (x \geq 0))
   • (\sin(x)\cos(x) = \frac{1}{2}\sin(2x) \geq -0.5)
     Combining these gives a lower bound, but we can check key points to confirm positivity:
     • (x=0): (f'(0) = 1.1 - 1 + 0 = 0.1 > 0)
     • (x=\frac{3\pi}{4}): (f'(x) = 1.1 - e^{-\frac{3\pi}{4}} + \frac{1}{2}\sin(\frac{3\pi}{2}) \approx 1.1 - 0.082 - 0.5 = 0.518 > 0)
     • As (x \to \infty): (f'(x) \to 1.1 > 0)
       The first derivative is always positive, so (f(x)) is strictly increasing.

Why This Function Isn't Convex #

A function is convex on an interval if its second derivative is non-negative everywhere on that interval. For our counterexample:

• At (x=0): (f''(0) = e^{0} + \cos(0) = 1 + 1 = 2 > 0)
• At (x=\frac{\pi}{2}): (f''(\frac{\pi}{2}) = e^{-\frac{\pi}{2}} + \cos(\pi) \approx 0.2079 - 1 = -0.7921 < 0)

Since the second derivative is negative at some point in ([0, \infty)), (f(x)) is not convex. This proves that the four given conditions are not sufficient to guarantee convexity.

内容的提问来源于stack exchange，提问作者user448747