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When I read about convex functions, I often encounter theorems of the following form: For a continuous function $f$ on the interval $[a;b]$ satisfying $f''(x) \geq 0$ for all $x \in [a;b]$, then $f$ is a convex function on $[a;b]$.

I think it is possible to relax the sufficient condition at both endpoints while maintaining continuity. Specifically:

For a continuous function $f$ on the interval $[a;b]$ satisfying $f''(x) \geq 0$ for all $x \in (a;b)$, prove that $f$ is a convex function on $[a;b]$.

I see that functions like $f(x)=-\sqrt{x}$ satisfy this condition. However, I don't yet know how to prove it. Thank you!

Martin R
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Math_fun2006
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1 Answers1

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If $f:[a, b] \to \Bbb R$ is twice differentiable on the open interval $(a, b)$ with $f''(x) \ge 0$ then $f$ is convex on $(a, b)$. This can be proven with Taylor's theorem, see for example Second derivative positive $\implies$ convex.

If $f$ is continuous on the closed interval $[a, b]$ and convex on $(a,b)$ then it is then $f$ is convex on $[a, b]$.

Proof: For $a < x < y < z < b$ is $$ \tag{$*$} f(y) \le \frac{z-y}{z-x} \cdot f(x) + \frac{y-x}{z-x} \cdot f(y) \, . $$ Since $f$ is continuous, we can take the limit $x \to a$ and/or $z \to b$, so that $(*)$ holds for $a \le x < y < z \le b$.

Remark: It can be shown that if $f:[a, b] \to \Bbb R$ is convex on $(a, b)$ then both limits $$ \lim_{x \to a} f(x) \, , \, \lim_{x \to b} f(x) $$ exist in $\Bbb R\cup \{ + \infty \}$. $f$ is convex on $[a, b]$ if and only if $$ \lim_{x \to a} f(x) \le f(a)\, , \, \lim_{x \to b} f(x) \le f(b) \, , $$ i.e. if $f$ is upper semicontinuous at $x=a$ and $x=b$.

Martin R
  • 128,226