# Hlawka-like Inequality for Convex Functions

### Lemma

If $a,b,c\,$ and $d\,$ are real numbers such that $0\le a\le b,c\le d\,$ and $a+d=b+c,\,$ then

$f(a)+f(d)\ge f(b)+f(c).$

### Proof of Lemma

Since $b\in [a,d],\,$ there is $k\in [0,1]\,$ such that $b=ka+(1-k)d.\,$ From $a+d=b+c\,$ we get $c=kd+(1-k)a.\,$ We have

$f(b)=f(ka+(1-k)d)\le kf(a)+(1-k)f(d)\\ f(c)=f(kd+(1-k)a)\le kf(d)+(1-k)f(a).$

Adding up those two, we get $f(a)+f(d)\ge f(b)+f(c).$

### Solution 1

Back to main problem now. We apply the lemma for the non-negative numbers $a=u,\,$ $b=u+y,\,$ $c=u+z,\,$ and $d=u+y+z,\,$ and obtain

(*)

\displaystyle \begin{align} &f(u)+f(u+y+z)-f(u+y)-f(u+z)\ge 0\,\Rightarrow\\ &\int_0^x[f(u)+f(u+y+z)-f(u+y)-f(u+z)]du\ge 0\,\Rightarrow\\ &\int_0^xf(u)du+\int_0^xf(u+y+z)du\ge\int_0^xf(u+y)du+\int_0^xf(u+z)du\,\Rightarrow\\ &g(x)+\int_0^xf(u+y+z)du\ge\int_0^xf(u+y)du+\int_0^xf(u+z)du. \end{align}

But

\displaystyle \begin{align} \int_0^xf(u+y+z)du &=\int_{y+z}^{x+y+z}f(u)du\\ &=\int_{0}^{x+y+z}f(u)du-\int_{0}^{y+z}f(u)du\\ &=g(x+y+z)-g(y+z). \end{align}

Similarly,

\displaystyle \begin{align} \int_0^{x}f(u+y)du &= g(x+y)-g(y),\\ \int_0^{x}f(u+z)du &= g(x+z)-g(z),\\ \end{align}

Substituting the last three equalities in (*) we get

$g(x)+g(x+y+z)-g(y+z)\ge g(x+y)-g(y)+g(x+z)-g(z)$

which is in fact the required inequality.

### Solution 2

Let $\displaystyle \phi(x)=\sum_{cycl}g(x)+g(x+y+z)-\sum_{cycl}g(x+y).\,$ Then

\begin{align}\phi'(x)&=g'(x)+g'(x+y+z)-g'(x+y)-g'(x+z)\\ &=f(x)+f(x+y+z)-f(x+y)-f(x+z). \end{align}

Now, $x\le x+y,x+z\le x+y+z,\,$ so that there are $\mu,\tau\in [0,1]\,$ such that

\begin{align} x+y&=\mu x+(1-\mu)(x+y+z)\\ x+z&=\tau x+(1-\tau)(x+y+z), \end{align}

implying

$2x+y+z=(\mu+\tau)x+(2-\mu-\tau)(x+y+z)$

which is the same as $(1-\mu-\tau)(y+z)=0.\,$ Then one of the two: $y+z\,$ implies $y=z=0\,$ and the required inequality reduces to $2g(x)+2g(0)\ge 2g(x)+g(0)\,$ which is true because $g(0)=0.\,$ If, on the other hand, $\mu+\tau=1,\,$ then, by the convexity of $f,$

\begin{align} f(x+y)&=f(\mu x+(1-\mu)(x+y+z))\le\mu f(x)+(1-\mu)f(x+y+z),\\ f(x+z)&=f(\tau x+(1-\tau)(x+y+z))\le\tau f(x)+(1-\tau)f(x+y+z) \end{align}

and, subsequently,

\begin{align} f(x+y)+f(x+z)&\le (\mu+\tau)f(x)+(2-\mu-\tau)(x+y+z))+f(x)\\ &=f(x)+f(x+y+z), \end{align}

i.e., $\phi'(x)\ge 0\,$ and so $\phi (x)\ge \phi(0)=0,\,$ which proves the required inequality.

### Solution 3

Let, WLOG, $z\ge y\ge x\,$ such that $y=x+\epsilon\,$ and $z=x+\epsilon+\xi,\,$ $\epsilon,\xi\ge 0.$

\displaystyle\begin{align} LHS&=2\int_{x+\epsilon}^{\xi+x+\epsilon}f(u)du+\int_{\xi+x+\epsilon}^{\xi+2x+\epsilon}f(u)du+\int_{\xi+2x+\epsilon}^{\xi+3x+2\epsilon}f(u)du\\ &\qquad\qquad\qquad+3\int_{x}^{x+\epsilon}f(u)du+4\int_{0}^{x}f(u)du. \end{align}

and

\displaystyle\begin{align} RHS&=2\int_{x+\epsilon}^{\xi+x+\epsilon}f(u)du+\int_{\xi+x+\epsilon}^{\xi+2x+\epsilon}f(u)du+\int_{\xi+x+\epsilon}^{\xi+2x+2\epsilon}f(u)du\\ &\qquad\qquad\qquad+3\int_{x}^{x+\epsilon}f(u)du+\int_{x+\epsilon}^{2x+\epsilon}f(u)du+3\int_{0}^{x}f(u)du. \end{align}

We need to prove that

$\displaystyle \int_{0}^{x}f(u)du+\int_{\xi+2x+\epsilon}^{3x+2\epsilon}f(u)du\ge\int_{x+\epsilon}^{2x+\epsilon}f(u)du+\int_{\xi+x+\epsilon}^{\xi+2(x+\epsilon)}f(u)du,$

which is satisfied owing to the convexity of $f\,$ as we have two exterior tranches (wings) vs two interior tranches ((inside wings), hard to explain to nonoption traders but it is an iron butterfly. The wings: $\displaystyle [x]\,|_{0}+[x+\epsilon]\,|_{\xi+2x+\epsilon}\,$ are greater than the inside wings,

$\displaystyle [x]\,|_{0}+[x+\epsilon]\,|_{\xi+2x+\epsilon}\ge[x]\,|_{x+\epsilon}+[x+\epsilon]\,|_{\xi+x+\epsilon}$

I used option language because this is exactly an Iron butterfly. More generally, for a convex function (a call price)

$C(K+\Delta)+C(K-\Delta)\ge 2C(K)\,\text{[butterfly]}$

and for $\Delta\gt\delta\gt 0,\,$

$C(K+\Delta)+C(K-\Delta)\ge C(K+\delta)+C(K-\delta)\,\text{[iron butterfly]}.$

And variations.

P.S. (AB): Here's the sequence of trades that implement the iron butterfly strategy:

1. Buy 1 OTM (out-of-money) Put
2. Sell 1 ATM (at-the-money) Put
3. Sell 1 ATM (at-the-money) Call
4. Buy 1 OTM (out-of-money) Call

### Acknowledgment

The problem which is due to Dan and Stefan Marinescu and Leo Giugiuc has been posted by Leo Giugiuc at the CutTheKnotMath facebook page. The problem was on a short list of the Romanian National Olympiad, grade 12. Solution 1 is by the authors, Solution 2 is by Diego Alvariz; Solution 3 is by N. N. Taleb (A generalization could be found in a separate file).

Note that the above generalizes the assertion of Marian Dinca's Hlawka-like inequality.