How to prove that the convexity of exponential function?
It is not allowed to use second derivative of $e^x$.
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1Possible duplicate of How to prove that $e^x$ is convex? – Dec 20 '15 at 07:12
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Second derivative is everywhere nonnegative. – dohmatob Dec 20 '15 at 11:25
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This means you prove: if $a<b$, then for all $t$ such that $0\le t\le 1$,
$$\begin{align*} f(ta+ (1-t)b) \leq tf(a) + (1-t)f(b)&\iff e^{ta+(1-t)b} \leq te^a+ (1-t)e^b\\ &\iff x^t\cdot y^{1-t} \leq tx+(1-t)y\\ &\iff r^t \leq tr + 1 - t\\ &\iff r^t - tr \leq 1 - t\\ &\iff t - tr \leq 1 - r^t\\ &\iff t \leq \dfrac{1-r^t}{1-r} \end{align*}$$
with $x = e^a$, $y = e^b$, $r = \dfrac{x}{y}$, $0 < x < y$, and $0 < r < 1$.
Brian M. Scott
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DeepSea
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Yeah I used it but how to prove this equality? I wanna prove it like we proved x². I mean in more detail. – zeynep Dec 20 '15 at 08:55
