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:


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

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?

  • $\begingroup$ 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

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.



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.


  • $\begingroup$ 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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