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Let $f(x)$ be a continuous and differentiable function on $\mathbb{R}$. Suppose $f(x)$ has a unique stationary point $a$, and that

$f''(a) \geq 0$

Now, ordinarily we cannot conclude that $a$ is a minimum since $f''(a)$ may be equal to $0$.

Suppose further though that we know $f''(x) \geq 0$ for all $x$ in the domain of $f(x)$ (as well as the fact $f'(a) = 0$, and $a$ is the unique stationary point).

Can we then safely conclude $f(a)$ is a minimum? My thinking is that since $f''(x) \geq 0$ for all $x$, the gradient can only be increasing in $x$, and as such inflection and maximum points are impossible, since both involve decreasing gradients in some way.

So that even though the second derivative is not strictly positive, we can still conclude a minimum?

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This is why convex functions are so amenable to analysis, local $\min$s are global $\min$s and first order conditions are sufficient.

If $f''\ge 0$ then $f$ is convex and then $a$ is a global $\min$ iff $f'(a) = 0$.

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Yes. If $f''(x)\geq 0$ for all $x$, then $f'$ is (nonstrictly) increasing on all of $\mathbb{R}$. Since $f'(a)=0$, this means $f'(x)\leq 0$ for all $x<a$ and $f'(x)\geq 0$ for all $x> a$. This means that if $x<a$ then $f(x)\geq f(a)$ and if $x>a$ then also $f(x)\geq f(a)$ (say, by the mean value theorem). Thus $f$ is minimized at $a$.

(With the assumption that $a$ is the unique stationary point, we can moreover conclude that $f(x)>f(a)$ strictly for all $x\neq a$, since if we had $f(x)=f(a)$ there would have to be a stationary point between them. In fact, since $f'$ is increasing, $f'$ would have to be identically $0$ on the entire interval between $a$ and $x$.)

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Proof is easy if we applying Taylor formula: $$ f(x) = f(a) + (x-a)f'(a) + (x-a)^2 \frac {f''(c)}2, \quad [c = tx + (1-t)a, t \in (0, 1)] $$ then $f'(a) = 0, f'(x) \geqslant 0 $ for all $x$ implies $$ f(x) = f(a) +\frac {f''(c)}2 (x-a)^2 \geqslant f(a), $$ then $f(a)$ is a minimum.

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