Differentiable Bounded Convex Real Function is Constant
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Theorem
Let $f$ be a real function which is:
- $(1): \quad$ differentiable on $\R$
- $(2): \quad$ bounded on $\R$
- $(3): \quad$ convex on $\R$.
Then $f$ is constant.
Proof
Let $f$ be differentiable and bounded on $\R$.
Let $f$ be convex on $\R$.
Let $\xi \in \R$.
Aiming for a contradiction, suppose $\map {f'} \xi > 0$.
Then by Mean Value of Convex Real Function it follows that:
- $\map f x \ge \map f \xi + \map {f'} \xi \paren {x - \xi} \to + \infty$ as $x \to +\infty$
and therefore is not bounded.
Similarly, suppose $\map {f'} \xi < 0$.
Then by Mean Value of Convex Real Function it follows that:
- $\map f x \ge \map f \xi + \map {f'} \xi \paren {x - \xi} \to + \infty$ as $x \to -\infty$
and therefore is likewise not bounded.
Hence $\map {f'} \xi = 0$.
From Zero Derivative implies Constant Function, it follows that $f$ is constant.
$\blacksquare$
Also see
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 12.21 \ (5)$