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I'm looking for a proof which shows that $$ \text{exp}(x) := \sum_{n = 0}^\infty \frac{x^n}{n!} $$ is convex, which means that for any $x,y\in\mathbb{R}$ and any $\lambda\in (0,1)$, one has $$ \text{exp}(\lambda x + (1-\lambda)y) \leq \lambda\text{exp}(x)+(1-\lambda)\text{exp}(y) $$

where it is not allowed to use any argument involving derivatives but only the definition above.

Thanks in advance.

aphorisme
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    This - https://math.stackexchange.com/questions/1367645/direct-proof-for-convexity-of-ex - will help. – rookie Jun 12 '17 at 07:22
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    "Not allowed", it's one of those stunts, again... I hope I am allowed to use both hands while typing. Usually, the result are "proofs" tacitly using other properties, as in the link @stud_iisc provided. It's next to impossible to derive any inequalities for $e^x$ from the series for $x<0$, you have to shift that into the region of positive arguments. The obvious means is $e^{x+y}=e^x e^y,$ and it's possible to prove that from the series, but it's not obvious how to do so in a rigorous way. –  Jun 12 '17 at 07:39
  • Thanks for the link @stud_iisc ; I agree, Professor Vector, point was, that this was an exercise in a lecture on analysis way before we had anything about derivatives. In fact, the way the exercise was stated, especially the points one gets from it suggests that there should be an easy way – since I haven't seen one so far, I thought to myself: maybe I've overseen something. But it seems: not I have overseen something. – aphorisme Jun 12 '17 at 09:25

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