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While reading a machine learning paper, I came across the following statement:

The function $\dfrac{f(x)}{x}$ is convex, where $$f(x) = \log\left(\sum_{i = 1}^m \exp\left(c_i x^2\right)\right),$$ with $c_1, \dots, c_m \geq 0$ and $x>0$.

I know that in general, the log-of-sum-of-exponentials is convex, but why does it remain convex when it is multiplied by $\dfrac{1}{x}$?

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    Experimentally, the claim that $g$ is convex does appear to be true My (now deleted) flawed answer was for the wrong function. – 2017-02-06

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Let me assume that $x \geq 0$ as otherwise the function is clearly not convex.

The perspective of the log-sum-exp function is convex, so $$g(x,y) = \begin{cases} y \log\left(\sum\exp\left(c_i\frac{x}{y}\right)\right) & \text{if } y \geq \frac{1}{x}, x \geq 0 \\ \infty &\text{otherwise}\end{cases}$$ is convex too. The function you are interested in is the partial minimization of $g$: $h(x) = \inf_y \{ g(x,y) \}$, which is convex (as partial minimizations of convex functions are convex).