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I have read about Logarithmic function. We can use the second-order condition to show that the $f(x)=\log_2(1+x), x \geq 0$ is a concave function. Now, is $g(x)$ a concave function? How can I prove this fact? \begin{equation} g(x)=(1-\frac{x}{a})\sum_{k=1}^K \log_2(1+\frac{b_kx}{-c_kx^2+d_kx+e_k}), \quad x \in [0,a] , \quad a>1 , \quad \forall k \in\{1,\ldots,K\}, \quad b_k,c_k,d_k,e_k>0 \end{equation} where $a,b_k,c_k,d_k,e_k$ are positive constants and $d_k$ , $e_k$ for all $k$ are enough large so that $-c_kx^2d_kx+e_k>0$.

Michael Hardy
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Hossein
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  • Please: Write $\log_2(1+x)$, not $log_2(1+x)$. It is coded as \log_2(1+x). The backslash not only prevents italicization but also provides proper spacing in things like $a\log b$ and $a\log(b)$ (notice that the latter has less space to the right of $\log$ than the former has). $\qquad$ – Michael Hardy Feb 20 '16 at 17:05
  • Convexity is the exception, rather than the rule. In my experience, If you can't figure out how to prove convexity, it's almost certainly not. – Michael Grant Feb 22 '16 at 14:49
  • Michael Hardy: Ok. Thanks for your guidance. – Hossein Feb 22 '16 at 19:28
  • Michael Grant: I think $g(x)$ is a concave function. For specific values of constants, the plot of the function confirms this fact. But I don't know the mathematical proof. – Hossein Feb 22 '16 at 19:53

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