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Undid revision 1207273350 by MohammadHoseinAkbari (talk) The 0 function is convex as well. Revert good faith edit |
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* If <math>f</math> is a convex function of one real variable, and <math>f(0)\le 0</math>, then <math>f</math> is [[Superadditivity|superadditive]] on the [[positive reals]], that is <math>f(a + b) \geq f(a) + f(b)</math> for positive real numbers <math>a</math> and <math>b</math>.
{{math proof|proof=
Since <math>f</math> is convex, by using one of the convex function definitions above and letting <math>x_2 = 0,</math> it follows that for all real <math>0 \leq t \leq 1,</math> <math display=block>
\begin{align} f(tx_1) & = f(t x_1 + (1-t) \cdot 0) \\ & \leq t f(x_1) + (1-t) f(0) \\ & \leq t f(x_1). \ \end{align} </math> From <math>f(tx_1)\leq t f(x_1) \begin{align} f(a) + f(b) & = f \left((a+b) \frac{a}{a+b} \right) + f \left((a+b) \frac{b}{a+b} \right) \\ & \leq \frac{a}{a+b} f(a+b) + \frac{b}{a+b} & = f(a+b).\\
\end{align}</math>
Namely, <math>f(a) + f(b) \leq f(a+b)</math>.
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