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


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