hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $f$ be concave on $I$. Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$. Then: {{begin-eqn}} {{eqn | l = \map f {\alpha x + \beta y}       | o = \ge       | r = \alpha \map f x + \beta \map f y       | c = {{Defof|Concave Real Function}} }} {{eqn | ll= \leadsto       | l = -\map f {\alpha x + \beta y}       | o = \le       | r = -\paren {\alpha \map f x + \beta \map f y}       | c =  }} {{eqn | r = \alpha \paren {-\map f x} + \beta \paren {-\map f y}       | c =  }} {{end-eqn}} and so $-f$ is convex by definition. {{qed|lemma}} === Sufficient Condition === Let $-f$ be convex on $I$. Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$. Then: {{begin-eqn}} {{eqn | l = -\map f {\alpha x + \beta y}       | o = \le       | r = \alpha \paren {-\map f x} + \beta \paren {-\map f y}       | c = {{Defof|Convex Real Function}} }} {{eqn | ll= \leadsto       | l = \map f {\alpha x + \beta y}       | o = \ge       | r = - \paren {\alpha \paren {-\map f x} + \beta \paren {-\map f y} }       | c =  }} {{eqn | r = \alpha \map f x + \beta \map f y       | c =  }} {{end-eqn}} and so $f$ is concave by definition. {{qed}}
Let $f$ be strictly convex real function on $I$ according to definition 1. That is: :$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \, \map f x + \beta \, \map f y$ Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$. As $\alpha + \beta = 1$, we have: :$x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$ Thus: :$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}$ :$\beta = \dfrac {x_2 - x_1} {x_3 - x_1}$ So: {{begin-eqn}} {{eqn | n = 1       | l = \map f {x_2}       | o = <       | r = \frac {x_3 - x_2} {x_3 - x_1} \, \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}       | c = substituting for $\alpha$ and $\beta$ }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = <       | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = <       | r = \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}       | o = <       | r = \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}       | o = <       | r = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}       | c =  }} {{end-eqn}} Again from $(1)$: {{begin-eqn}} {{eqn | n = 1       | l = \map f {x_2}       | o = <       | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = <       | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_1} \map f {x_2}       | o = <       | r = \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}       | o = <       | r = \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}       | o = <       | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}       | c =  }} {{end-eqn}} Thus: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ So: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$ demonstrating that $f$ is strictly convex on $I$ according to definition 3, and: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ demonstrating that $f$ is strictly convex on $I$ according to definition 2. As each step is an equivalence, the argument reverses throughout. {{qed}} Category:Convex Real Functions iv7r4z0j5dr5r9p1shuq2fus954kxnz
Let $f$ be concave on $\left({a \,.\,.\, b}\right)$. Then by definition of concavity: :$\forall x_1, x_2, x_3 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$ Let $0 < h_1 < h_2$. Substitute $x_1 = x$, $x_2 = x + h_1$, $x_3 = x + h_2$. Then: :$\dfrac {f \left({x + h_1}\right) - f \left({x}\right)} {h_1} \ge \dfrac {f \left({x + h_2}\right) - f \left({x}\right)} {h_2}$ Hence the function $F \left({h}\right) = \dfrac {f \left({x + h}\right) - f \left({x}\right)} h$ decreases in some $\left({0 \,.\,.\, \delta}\right)$. Thus from Limit of Decreasing Function it follows that both $\displaystyle \lim_{h \to 0^+} F \left({h}\right) = f'_+ \left({x}\right)$ and $\displaystyle \lim_{h \to 0^-} F \left({h}\right) = f'_-\left({x}\right)$ exist. {{qed}}
From Concave Real Function is Left-Hand and Right-Hand Differentiable, we have: :$\displaystyle \lim_{h \to 0^-} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^-} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^-} h}\right) = 0$ and similarly: :$\displaystyle \lim_{h \to 0^+} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^+} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^+} h}\right) = 0$ {{qed}}
Let $x, y \in \R$. Let $\alpha, \beta \in \R_{\ge 0}$ where $\alpha + \beta = 1$. {{begin-eqn}} {{eqn | l = \map f {\alpha x + \beta y}       | r = \size {\alpha x + \beta y}       | c = Definition of $f$ }} {{eqn | o = \le       | r = \size {\alpha x} + \size {\beta y}       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \size \alpha \size x + \size \beta \size y       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | r = \alpha \size x + \beta \size y       | c = {{Defof|Absolute Value}} }} {{eqn | r = \alpha \, \map f x + \beta \, \map f y       | c = Definition of $f$ }} {{end-eqn}} Hence the result by definition of convex real function. {{qed}}
=== Necessary Condition === Let $f$ be concave on $\openint a b$. Let $x_1, x_2, x_3, x_4 \in \openint a b$ such that: :$x_1 < x_2 < x_3 < x_4$ By the definition of concave: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} \ge \dfrac {\map f {x_4} - \map f {x_3} } {x_4 - x_3}$ Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$. Thus: :$\map {f'} {x_1} \ge \map {f'} {x_4}$ Hence $f'$ is decreasing on $\openint a b$. {{qed|lemma}} === Sufficient Condition === Let $f'$ be decreasing on $\openint a b$. Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$. By the Mean Value Theorem: :$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$ :$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$ where $x_1 < \xi < x_2 < \eta < x_3$. Since $f'$ is decreasing: :$\map {f'} \xi \ge \map {f'} \eta$ Thus: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Hence $f$ is concave by definition. {{qed}}
=== Necessary Condition === Let $f$ be strictly convex on $\openint a b$. Let $r, s \in \openint a b$ be arbitrarily selected such that $r < s$. We are to show that: :$\map {f'} r < \map {f'} s$ Let $x_1, x_2, x_3 \in \openint a b$ be chosen such that: :$r < x_1 < x_2 < x_3 < s$ By the definition of strictly convex: :$(1): \quad \dfrac {\map f {x_1} - \map f r} {x_1 - r} < \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} < \dfrac {\map f s - \map f {x_3} } {s - x_3}$ and: :$(2): \quad \dfrac {\map f {x_2} - \map f r} {x_2 - r} < \dfrac {\map f s - \map f {x_2} } {s - x_2}$ Let $x_1 \to r^+$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\map f {x_1} - \map f r} {x_1 - r}       | o = <       | r = \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_1} - \map f r} {x_1 - r}       | o = \le       | r = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}       | c = Limits Preserve Inequalities }} {{eqn | n = 3       | ll= \leadsto       | l = \map {f'} r       | o = \le       | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}       | c = {{Defof|Derivative of Real Function at Point}} }} {{end-eqn}} Similarly, let $x_3 \to s^-$. We have: {{begin-eqn}} {{eqn | l = \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}       | o = <       | r = \dfrac {\map f s - \map f {x_3} } {s - x_3}       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \lim_{x_3 \mathop \to s^-} \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}       | o = \le       | r = \lim_{x_3 \mathop \to s^-} \dfrac {\map f s - \map f {x_3} } {s - x_3}       | c = Limits Preserve Inequalities }} {{eqn | n = 4       | ll= \leadsto       | l = \dfrac {\map f s - \map f {x_2} } {s - x_2}       | o = \le       | r = \map {f'} s       | c = {{Defof|Derivative of Real Function at Point}} }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = \map {f'} r       | o = \le       | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}       | c = from $(3)$ }} {{eqn | o = <       | r = \dfrac {\map f s - \map f {x_2} } {s - x_2}       | c = from $(2)$ }} {{eqn | o = \le       | r = \map {f'} s       | c = from $(4)$ }} {{end-eqn}} Thus: :$\map {f'} r < \map {f'} s$ Hence $f'$ is strictly decreasing on $\openint a b$. {{Proofread}} {{qed|lemma}} === Sufficient Condition === Let $f'$ be strictly increasing on $\openint a b$. Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$. By the Mean Value Theorem: :$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$ :$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$ where $x_1 < \xi < x_2 < \eta < x_3$. Since $f'$ is strictly increasing: :$\map {f'} \xi < \map {f'} \eta$ Thus: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Hence $f$ is strictly convex by definition. {{qed}}
=== Necessary Condition === Let $f$ be strictly concave on $\openint a b$. Let $r, s \in \openint a b$ be arbitrarily selected such that $r < s$. We are to show that: :$\map {f'} r > \map {f'} s$ Let $x_1, x_2, x_3 \in \openint a b$ be chosen such that: :$r < x_1 < x_2 < x_3 < s$ By the definition of strictly concave: :$(1): \quad \dfrac {\map f {x_1} - \map f r} {x_1 - r} > \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} > \dfrac {\map f s - \map f {x_3} } {s - x_3}$ and: :$(2): \quad \dfrac {\map f {x_2} - \map f r} {x_2 - r} > \dfrac {\map f s - \map f {x_2} } {s - x_2}$ Let $x_1 \to r^+$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\map f {x_1} - \map f r} {x_1 - r}       | o = >       | r = \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_1} - \map f r} {x_1 - r}       | o = \ge       | r = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}       | c = Limits Preserve Inequalities }} {{eqn | n = 3       | ll= \leadsto       | l = \map {f'} r       | o = \ge       | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}       | c = {{Defof|Derivative of Real Function at Point}} }} {{end-eqn}} Similarly, let $x_3 \to s^-$. We have: {{begin-eqn}} {{eqn | l = \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}       | o = >       | r = \dfrac {\map f s - \map f {x_3} } {s - x_3}       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = \lim_{x_3 \mathop \to s^-} \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}       | o = \ge       | r = \lim_{x_3 \mathop \to s^-} \dfrac {\map f s - \map f {x_3} } {s - x_3}       | c = Limits Preserve Inequalities }} {{eqn | n = 4       | ll= \leadsto       | l = \dfrac {\map f s - \map f {x_2} } {s - x_2}       | o = \ge       | r = \map {f'} s       | c = {{Defof|Derivative of Real Function at Point}} }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = \map {f'} r       | o = \ge       | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}       | c = from $(3)$ }} {{eqn | o = >       | r = \dfrac {\map f s - \map f {x_2} } {s - x_2}       | c = from $(2)$ }} {{eqn | o = \ge       | r = \map {f'} s       | c = from $(4)$ }} {{end-eqn}} Thus: :$\map {f'} r > \map {f'} s$ Hence $f'$ is strictly decreasing on $\openint a b$. {{Proofread}} {{qed|lemma}} === Sufficient Condition === Let $f'$ be strictly decreasing on $\openint a b$. Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$. By the Mean Value Theorem: :$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$ :$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$ where $x_1 < \xi < x_2 < \eta < x_3$. Since $f'$ is strictly decreasing: :$\map {f'} \xi > \map {f'} \eta$ Thus: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Hence $f$ is strictly concave by definition. {{qed}}
Let: :$X = f \left({x}\right) \in J$ :$Y = f \left({y}\right) \in J$. From the definition of convex: : $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$ Let $f$ be strictly decreasing on $I$. Then from Inverse of Strictly Monotone Function it follows that $f^{-1}$ is strictly decreasing on $J$. Thus: : $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \le f^{-1} \left({\alpha X + \beta Y}\right)$ Hence $f^{-1}$ is convex on $J$. {{qed}}
Let $X = \map f x \in J, Y = \map f y \in J$. From the definition of concave: :$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \ge \alpha \map f x + \beta \map f y$ Let $f$ be strictly increasing on $I$. From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly increasing on $J$. Thus: :$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y \ge \map {f^{-1} } {\alpha X + \beta Y}$ Hence $f^{-1}$ is convex on $J$. {{qed}}
Let: :$X = f \left({x}\right) \in J$ :$Y = f \left({y}\right) \in J$. From the definition of concave: : $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \ge \alpha f \left({x}\right) + \beta f \left({y}\right)$ Let $f$ be strictly decreasing on $I$. From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly decreasing on $J$. Thus: : $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \ge f^{-1} \left({\alpha X + \beta Y}\right)$ Hence $f^{-1}$ is concave on $J$. {{qed}}
By the Mean Value Theorem: : $\displaystyle \exists \eta \in \left({x \,.\,.\, \xi}\right): f^{\prime} \left({\eta}\right) = \dfrac {f \left({x}\right) - f \left({\xi}\right)} {x - \xi}$ From Real Function is Concave iff Derivative is Decreasing, the derivative of $f$ is decreasing. Thus: : $x > \xi \implies f' \left({\eta}\right) \le f' \left({\xi}\right)$ : $x < \xi \implies f' \left({\eta}\right) \ge f' \left({\xi}\right)$ Hence: : $f \left({x}\right) - f \left({\xi}\right) \le f' \left({\xi}\right) \left({x - \xi}\right)$ {{qed}}
Let $f$ be differentiable and bounded on $\R$. Let $f$ be concave on $\R$. Let $\xi \in \R$. Aiming for a contradiction, suppose $f' \left({\xi}\right) > 0$. Then by Mean Value of Concave Real Function it follows that: : $f \left({x}\right) \le f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to -\infty$ as $x \to +\infty$ and therefore is not bounded. Similarly, suppose $f' \left({\xi}\right) < 0$. Then by Mean Value of Concave Real Function it follows that: : $f \left({x}\right) \le f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to -\infty$ as $x \to -\infty$ and therefore is likewise not bounded. Hence $f' \left({\xi}\right) = 0$. From Zero Derivative implies Constant Function, it follows that $f$ is constant. {{qed}}
From Real Function is Strictly Concave iff Derivative is Strictly Decreasing, $f$ is strictly concave {{iff}} $f'$ is strictly decreasing. From Derivative of Monotone Function, $f'$ is strictly decreasing {{iff}} its second derivative $f'' < 0$. {{qed}}
From Real Function is Strictly Convex iff Derivative is Strictly Increasing, $f$ is strictly convex {{iff}} $f'$ is strictly increasing. From Derivative of Monotone Function, $f'$ is strictly increasing {{iff}} its second derivative $f'' > 0$. {{qed}}
Let: :$X = \map f x \in J$ :$Y = \map f y \in J$. From the definition of strictly convex: :$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \map f x + \beta \map f y$ Let $f$ be strictly decreasing on $I$. Then from Inverse of Strictly Monotone Function it follows that $f^{-1}$ is strictly decreasing on $J$. Thus: :$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y < \map {f^{-1} } {\alpha X + \beta Y}$ Hence $f^{-1}$ is strictly convex on $J$. {{qed}}
Let: :$X = f \left({x}\right) \in J$ :$Y = f \left({y}\right) \in J$. From the definition of strictly concave: : $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) > \alpha f \left({x}\right) + \beta f \left({y}\right)$ Let $f$ be strictly decreasing on $I$. From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly decreasing on $J$. Thus: : $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y > f^{-1} \left({\alpha X + \beta Y}\right)$ Hence $f^{-1}$ is strictly concave on $J$. {{qed}}
Let: :$X = \map f x \in J$ :$Y = \map f y \in J$ From the definition of strictly convex: :$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \, \map f x + \beta \, \map f y$ Let $f$ be strictly increasing on $I$. From Inverse of Strictly Monotone Function it follows that, $f^{-1}$ is strictly increasing on $J$. Thus: :$\alpha \, \map {f^{-1} } X + \beta \, \map {f^{-1} } Y = \alpha x + \beta y < \map {f^{-1} } {\alpha X + \beta Y}$ Hence $f^{-1}$ is strictly concave on $J$. {{qed}}
Let $X = f \left({x}\right) \in J, Y = f \left({y}\right) \in J$. From the definition of Strictly concave: : $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) > \alpha f \left({x}\right) + \beta f \left({y}\right)$ Let $f$ be strictly increasing on $I$. From Inverse of Strictly Monotone Function, it follows that $f^{-1}$ is strictly increasing on $J$. Thus: : $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y > f^{-1} \left({\alpha X + \beta Y}\right)$ Hence $f^{-1}$ is strictly convex on $J$. {{qed}}
By definition of the Gamma function: :$\displaystyle \map \Gamma x = \int_0^{\to \infty} t^{x - 1} e^{-t} \rd t$ In order to allow the limits to be evaluated, this is to be expressed as: :$\displaystyle \map \Gamma x = \lim_{\delta \mathop \to 0^+, \ \Delta \mathop \to +\infty} \int_\delta^\Delta t^{x - 1} e^{-t} \rd t$ where $0 < \delta < \Delta$. Let $-t = \ln u$. Then by Derivative of Logarithm Function and the Chain Rule for Derivatives: :$-1 = \dfrac 1 u \dfrac {\rd u} {\rd t}$ Also: {{begin-eqn}} {{eqn | l = -t       | r = \ln u }} {{eqn | ll= \leadsto       | l = e^{-t}       | r = u       | c = Exponential of Natural Logarithm }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \int_\delta^\Delta t^{x - 1} e^{-t} \rd t       | r = \int_{e^{-\delta} }^{e^{-\Delta} } \paren {-\ln u}^{x - 1} e^{\ln u} \paren {-\frac 1 u \frac {\rd u} {\rd t} } \rd t       | c = Integration by Substitution }} {{eqn | r = \int_{e^{-\Delta} }^{e^{-\delta} } \paren {-\ln u}^{x - 1} e^{\ln u} \paren {\frac 1 u \frac {\rd u} {\rd t} } \rd t       | c = Reversal of Limits of Definite Integral }} {{eqn | r = \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} u \frac 1 u \rd u       | c = Exponential of Natural Logarithm }} {{eqn | n = 1       | r = \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} \rd u       | c = simplification }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = \lim_{\delta \mathop \to 0^+} e^{-\delta}       | r = 1       | c = Exponential of Zero }} {{eqn | l = \lim_{\Delta \mathop \to +\infty} e^{-\Delta}       | r = 0       | c = Exponential Tends to Zero and Infinity }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \lim_{\delta \mathop \to 0^+, \ \Delta \mathop \to +\infty} \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} \rd u       | r = \int_{e^{-\Delta} }^1 \paren {\ln \frac 1 u}^{x - 1} \rd u       | c = substituting for the top limit }} {{eqn | r = \int_{\to 0}^1 \paren {\ln \frac 1 u}^{x - 1} \rd u       | c = as $\ln 0$ is undefined }} {{end-eqn}} The result follows by renaming the dummy variable $u$ to $t$. {{qed}}
{{begin-eqn}} {{eqn | l = \int_a^b \map f x \rd x + \int_b^a \map f x \rd x       | r = \int_a^a \map f x \rd x       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = 0       | c = Definite Integral on Zero Interval }} {{eqn | ll = \leadsto       | l = \int_a^b \map f x \rd x       | r = - \int_b^a \map f x \rd x }} {{end-eqn}} {{CircularStructure|Careful with the above. Down at the bottom of Sum of Integrals on Adjacent Intervals for Integrable Functions use is made of this result. There has been a lot of activity on this area of mathematics, and several people have contributed, possibly without having completely followed through all the chains of implications. </br> My bad. Either way this has to be proven independently since the case $a {{=}} b$ was not considered in that theorem.}}
Consider the following inequalities, valid for $0 < t < 1$: {{begin-eqn}} {{eqn | l = t^{x - 1} \left({1 - t}\right)^{y - 1}       | o = <       | r = t^{x - 1}       | c =  }} {{eqn | ll= \implies       | l = t^{x - 1} \left({1 - t}\right)^{y - 1}       | o = <       | r = \left({1 - t}\right)^{y - 1}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \ \mathrm d t       | r = \left.{\frac {t^x} x}\right\vert_{\mathop \to 0}^{\mathop \to 1}       | c =  }} {{eqn | r = x - 0       | c =  }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = \int_{\mathop \to 0}^{\mathop \to 1} \left({1 - t}\right)^{y - 1} \ \mathrm d t       | r = -\left.{\frac {\left({1 - t}\right)^y} y}\right\vert_{\mathop \to 0}^{\mathop \to 1}       | c =  }} {{eqn | r = y - 0       | c =  }} {{end-eqn}} The result follows from the Comparison Test for Improper Integral. {{qed}}
{{WLOG}}, we consider the case $I = \openint 0 \to$ such that $\displaystyle l = \int_0^{\mathop \to +\infty} \map \phi x \rd x$ exists. Let: :$\ds a_n = \int_{n - 1}^n \map f x \rd x$ :$\ds b_n = \int_{n - 1}^n \map \phi x \rd x$ for $n = 1, 2, \ldots$ Then the series: :$\ds \sum_{n \mathop = 1}^\infty b_n$ is a convergent series of positive terms whose sum is $l$. Also: $\size {a_n} \le b_n$ for $n = 1, 2, \ldots$ Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ a convergent series by the Comparison Test. That is: :$\ds m = \lim_{N \mathop \to \infty} \int_0^N \map f x \rd x$ exists. Given any $X > 0$, $N$ can be taken to be the smallest natural number which satisfies $N > X$. Then: {{begin-eqn}} {{eqn | l = \size {\int_1^X \map f x \rd x - \int_1^{N - 1} \map f x \rd x}       | o = \le       | r = \int_{N - 1}^X \size {\map f x} \rd x       | c =  }} {{eqn | o = \le       | r = b_N       | c =  }} {{eqn | o = \to       | r = 0 \text{ as } N \to \infty       | c =  }} {{end-eqn}} It follows that: :$\ds \lim_{X \mathop \to \infty} \int_1^X \map f x \rd x = m$ as required. The other cases are treated similarly. {{qed}}