# Find function such that $\sum y_i\le\sum x_i\Rightarrow\sum f(y_i)\le\sum f(x_i)$

What kind of a function $f$ must be to satisfy the following?

If $\sum_{i=1}^{n} y_i \leq \sum_{j=1}^{n} x_j$, where $x_j, y_i \in [0,1],\forall i,j$ then $$\sum_{i=1}^{n} f(y_i) \leq \sum_{j=1}^{n} f(x_j).$$

Any help would be appreciated. Thanks in advance!

Preferably $f$ must be convex and increasing.

$f$ is linear from the answer given by the user grand_chat. What if the above inequalities($\leq$) are replaced by the strict inequality ($<$)?

The only solutions are linear, i.e. $f(x) = cx + b$.

Wlog we can assume $f(0)=0$ by considering $g(x):=f(x)-f(0)$. Also, $f$ is nondecreasing; this follows from applying the stated property with $n=1$.

Taking $x_1+x_2 = y_1+y_2$ and applying the stated property twice (in both directions of inequality) we obtain: $$f(y_1)+f(y_2) = f(x_1) + f(x_2)\qquad\text{whenever y_1+y_2=x_1+x_2.}\tag1$$

Finally note that $(x+y) + 0 = x + y$ so (1) gives $$f(x+y) = f(x) + f(y).\tag2$$ Equation (2) is the famous Cauchy functional equation. Property (2) along with the monotonicity of $f$ implies that $f(x)=cx$ for some $c\ge0$ (since $f$ is nondecreasing).

EDIT: If we relax the inequality $(\le)$ to a strict inequality $(<)$, the same result follows except we rule out the possibility $c=0$. This follows from a continuity argument:

1. First show that (2) holds whenever $y$ is a continuity point for $f$, by considering $$x+(y-\epsilon) < (x+y) + 0 < x + (y+\epsilon).$$

2. $f$ is monotone, so has only countably many discontinuities. Let $x_0$ be a continuity point for $f$, and let $\{ y_n \}$ be a sequence of continuity points tending to $0$. We have for all $n$ $$f(y_n) = f(x_0+y_n) - f(x_0).$$ Since $f$ is continuous at $x_0$, we have $\lim f(y_n)=0$. Monotonicity of $f$ forces $f$ to be continuous at zero.

3. Lastly, let $x$ be arbitrary and let $\{ y_n\}$ be a sequence of continuity points tending to zero. Letting $n\to\infty$ in the identity $$f(x+y_n)=f(x) + f(y_n)$$ shows that $f$ is continuous at $x$. Hence every point is a continuity point for $f$, and therefore (2) holds everywhere.

• Agreed. What if I have $<$ instead of $\leq$ in both the hypothesis and the implication property? Thanks for the earlier answer! – rookie May 10 '17 at 5:29
• Shouldn't $c$ be a non-negative real number? – PN Karthik May 10 '17 at 5:52
• And a small error. Since $f$ is nondecreasing, $c$ must be positive. – rookie May 10 '17 at 6:16
• @Karthik Yes, $c\ge0$. I've made that correction. – grand_chat May 10 '17 at 13:46
• @stud_iisc Nope. For example, $f(x):=x^2$ is strictly convex but it is not true that $0^2 + 3^2 < 2^2 + 2^2$. – grand_chat May 10 '17 at 15:25