Prove that the sum of a convex function and a concave function is convex.

Question

I was trying to prove, for the sake of curiosity, if the sum of a convex and a concave functions is convex, so i tried to do the following:

Let f: R -> R and g: R -> R.
f is convex and g is concave.
Prove that (f + g) : R -> R is convex.

Proof:
Since f is convex, by definition holds that:
$f(tx + (1-t)y) < tf(x) + (1-t)f(y), \forall x,y \in \Re, t \in (0,1)$
And g is concave, so:
$g(tx + (1-t)y) > tg(x) + (1-t)g(y), \forall x,y \in \Re, t \in (0,1)$

Now, I can write the inequalities like:
$f(tx + (1-t)y) - tf(x) - (1-t)f(y)< 0$
$-g(tx + (1-t)y) + tg(x) + (1-t)g(y) < 0$

It follows that: $f(tx + (1-t)y) - tf(x) - (1-t)f(y)< tg(x) + (1-t)g(y) - g(tx + (1-t)y)$
Doing some calculation, I came up with:
$f(tx + (1-t)y) + g(tx + (1-t)y) < tg(x) + (1-t)g(y) + tf(x) + (1-t)f(y)$

I don't know if I have made some reasoning or logical mistakes, in fact it looks fine to me, but I don't trust myself so much to say that it's correct, so please, let me know. Thanks.

Answer 1

It's obviously not true. If that is true then you can choose $f_1(x) = -g(x)$ and $g_1(x) = -f(x)$, then $f_1(x)$ is convex and $g_1(x)$ is concave, and $f_1(x)+g_1(x)$ = $-(f(x)+g(x))$ is concave, hence contradiction.

Answer 2

Your statement after "It follows that:" is false. You are essentially claiming that if $A<0$ and $B<0$, then $A<B$, which is clearly false.

Answer 3

Your mistake is when you write

Since f is convex, by definition holds that:
$f(tx + (1-t)y) < tf(x) + (1-t)f(y), \forall x,y \in \Re, t \in (0,1)$
And g is concave, so:
$g(tx + (1-t)y) < tg(x) + (1-t)g(y), \forall x,y \in \Re, t \in (0,1)$

you put the same inequality sign (<), though they must be different. So you just took two functions of the same concavity.

This statement that you try to prove is obviously wrong, since concave and convex are “equal”, symmetric notions. Why would the sum of two functions of different concavity prefer one over another?