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Take two real differentiable convex functions, $f_1$ and $f_2$, defined on the unit interval $[0; 1]$. I want to find the global optimum of: $\min_{x \in [0;1]} af_1(x)+bf_2(x)$, for given $a, b \in \mathbb{R}$. Is there a simple solution to this?

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It's probably easier to answer if a and b are nonnegative, since $af_1(x)+bf_2(x)$ is convex when $a>0$ and $b>0$. –  Snowball Oct 5 '12 at 22:39
    
Just to add to Snowball's answer more explicitly, if a or b are negative then convexity is not guaranteed so in general it will be difficult to optimize. –  Bitwise Oct 5 '12 at 22:56
    
Moreover, if $a$ and $b$ are nonnegative and $f_1$ and $f_2$ have their global minima at $s$ and $t$, then $f = a f_1 + b f_2$ has a global minimum somewhere in the interval $[\min(s,t),\max(s,t)]$. –  Robert Israel Oct 5 '12 at 23:12
    
To expand on @Bitwise's comment: any $C^2$ function on $[0,1]$ can be written as the difference of two $C^2$ convex functions. –  Robert Israel Oct 5 '12 at 23:17
    
@RobertIsrael thanks Robert, I wasn't aware of that. Can you provide a reference? –  Bitwise Oct 6 '12 at 0:13
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