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I'm looking at the Boyd & Vandenberghe slides on Convex Optimization. In slide 18, it applies the rules of vector composition on an example to say that it is convex. The example given is

$\log\sum_{i=1}^{m}\exp g_i$ is convex if $g_i$ is convex.

I know the exponential (and sum of exponential) function is convex. But isn't the log function concave? How does this fit the rule, since $h$ (I'm assuming $h$ is the log function) isn't convex?

Rodrigo de Azevedo
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Rayne
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  • Related: http://math.stackexchange.com/questions/665768/how-to-prove-that-the-sum-of-two-log-convex-functions-is-log-convex – dxiv Feb 05 '16 at 17:40

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A function $f$ is called log-convex if $\ln f$ is convex. It is not that difficult to show that a sum of two log-convex functions is log-convex. All you need to do is to notice that the function $\exp g_i$ is log-convex.

Another approach would be to show by definition for the case $m=2$ and then generalise to an arbitrary $m$.

TZakrevskiy
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  • So in this case, $f(x) = \sum_{i=1}^{m}\exp g_i$, where $g(x) = g_i$, $h = \sum_{i=1}^{m}\exp(...)$, and $h$ (and therefore, $f$) is log-convex? The rule talks about $f$ being convex. What is the difference between $f$ being log-convex and $f$ being convex? – Rayne Feb 05 '16 at 18:12
  • Log-convexity implies convexity, but there are convex functions that are not log-convex ($x^2$, for example). Here you have convex functions $g_i$, hence the functions $\exp g_i$ are, by definition, log-convex, therefore, $\sum_i \exp g_i$ is log-convex, which, by definition, means that $\log \sum_i \exp g_i$ is convex. – TZakrevskiy Feb 05 '16 at 19:30