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As above, how would one mathematically prove that a linear combination of convex functions is also convex?

We know a function defined on a convex set $S$ is convex if:

$$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2)$$

where $t$ is from $0$ to $1$

We must prove that $\sum_{i=1}^n a_i f_i(x)$ is also convex given a bunch of functions $f_1, f_2$ etc.


How do i approach this problem? I could say the following:

$tf(x_1)+(1-t)f(x_2)+tg(x_1)+(1-t)g(x_2)=t(f(x_1)+g(x_1))+(1-t)(f(x_2)+g(x_2))$

$f(tx_1+(1-t)x_2)+g(tx_1+(1-t)x_2)\leq t(f(x_1)+g(x_1))+(1-t)(f(x_2)+g(x_2))$

Is this how we show it?


Secondly, how would we prove the same thing for a concave function? Isn't it just adding a - sign? How would i mathematically prove it?

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  • $\begingroup$ As for the concavity: how did you define concave functions? Why would you want to approach it differently in any way than convex functions? $\endgroup$ – Qi Zhu Sep 22 '18 at 21:02
  • $\begingroup$ Are the coefficients $a_i$ all non-negative? $\endgroup$ – robjohn Sep 22 '18 at 22:05
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In general your statement is false. The function $f(x)=x^2$ is convex, while $-f$ is not convex. It is true if you consider so-called conical combinations, i.e. all coefficients are supposed to be nonnegative.

The proof is trivial bacause it is enough to multiply the inequalities by nonnegative scalars and sum them up.

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hint

You should assume the coefficients positive. Your sum is finite, so you just need to prove that $$a_1f_1+a_2f_2$$ is convexe.

$$a_i>0.$$

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  • $\begingroup$ This is obviously false. Inequalities are not preserved by linear combinations. $\endgroup$ – Kabo Murphy Sep 22 '18 at 23:37
  • $\begingroup$ @KaviRamaMurthy It is true if the $a_i$s are positives. $\endgroup$ – hamam_Abdallah Sep 23 '18 at 8:59

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