# A beautiful inequality for convex functions

Let $f\in \mathcal{C}([0,1],\mathbb R_+)$ increasing.

Prove that there exist $g,h\in \mathcal{C}([0,1],\mathbb R)$, convexs, such that $g\leqslant f \leqslant h$ and : $$\displaystyle \frac{1}{2}\int_0^1 h \leqslant \int_0^1 f \leqslant 2\int_0^1 g$$

Let $H$ the Heaviside function : $f\leq g$

$=1$ for $x\geq0$ and $0$ for $x<0$ and $V(x) = x H(x) = \max(x,0)$.

Let $f$ increasing positive, let $N\geq1$ and $\delta_j = f(\frac jN) - f(\frac{j-1}N)$ for $1\leq j \leq N$,

Thus, $$f_N(x) = f(0) + \delta_1 H\Big(x - \frac 1 N\Big) + \delta_2 H\Big(x - \frac 2 N\Big) + \dotsb + \delta_{N-1} H\Big(x - \frac{N-1}N\Big)$$

The function $f_N$ is bounded above by $f$ (winch is increasing positive).

And, $$\int_0^1 f(u)\,du - \int_0^1 f_N(u)\,du \leq \frac1N(f(1) - f(0))$$

Now, define : $$\begin{split} g_N(x) = f(0) + \delta_1\Big(1-\frac 1 N\Big)^{-1} V\Big(x- \frac 1 N\Big) &+ \delta_2\Big(1-\frac 2 N\Big)^{-1} V\Big(x-\frac 2 N\Big) \\ &+ \dotsb + \delta_{N-1} N V\Big(x- \frac{N-1}N\Big)\end{split}$$

Therefore, $g_N$ is convex, and $g_N(x) \leq f_N(x)$.

Then : $$\forall x, \quad g_N(x) \leq f(x)$$

\begin{align*} \int_0^1 g_N(u)\,du &= f(0) + \frac 1 2\delta_1 \Big(1 - \frac 1 N\Big) + \frac 1 2\delta_2 \Big(1 - \frac 2 N\Big) + \dotsb +\frac 1 2 \delta_{N-1} \frac 1 N\\ \int_0^1 f_N(u)\,du &= f(0) + \delta_1 \Big(1 - \frac 1 N\Big) + \delta_2 \Big(1 - \frac 2 N\Big) + \dotsb +\delta_{N-1} \frac 1 N \\ \end{align*}

Therefore,

\begin{align*}\int_0^1 f(u)\,du &\leq 2(\int_0^1 g_N(u)\,du - f(0)) + f(0) + \frac1N\big(f(1) - f(0)\big)\\ \int_0^1 g_N(u)\,du &\geq \frac 1 2\int_0^1 f(u)\,du + \frac 1 2 f(0)-\frac 1{2N}(f(1)-f(0)) \end{align*}.

Then $\int_0^1 f \leqslant 2\int_0^1 g$.

How can I do for the second inequality ?

• Why do you have $g_N \leq f$ ? – user10676 Mar 7 '14 at 17:47
• @user10676 Look carefully my solution : $f_N$ is bounded above by $f$ and $g_n\leq f_N$ – user119228 Mar 7 '14 at 21:59
• Here is a closely related question. – 23rd Mar 9 '14 at 6:35
• @23rd it's a beautiful answer, can we proceed in the same way to prove the other bound ? – user119228 Mar 9 '14 at 12:29
• @Julien: I almost forgot all the details in my answer to the linked question, and I just looked back at it briefly. It seems that the same argument there does not work, because it uses the fact that the supremum of convex functions is still convex, which fails to be true for infimum in general. However, I think the method in the remark of my answer there(or the first version of my answer) still works for constructing $h$, and this method shares the same basic idea with your own approach. – 23rd Mar 9 '14 at 13:30

The statement holds for any increasing $f:[0,1]\to[0,+\infty)$, i.e. the continuity of $f$ is not necessary. Note that if the statement holds for some $f$, then after adding a common positive constant to $f,g,h$ simultaneously, the statement still holds, so for simplicity, we may assume $f(0)=0$.

Following your notations, define $H$ by letting $H(x)= 1$ when $x\ge 0$ and $H(x)=0$ when $x<0$. Define $V(x):=xH(x)$. Let us start with the simplest case $f(x)=f_t(x):=H(x-t)$ for some $t\in (0,1)$ and construct the associated $g_t$ and $h_t$. Functions of the form $x\mapsto b\cdot V(x-a)$ for appropriate constants $a< 1$ and $b>0$ are natural candidates for both $g_t$ and $h_t$, which are non-negative, continuous, increasing and convex. Using the constraints $g_t\le f_t\le h_t$ and $\frac{1}{2}\int_0^1 h_t\le\int_0^1 f_t\le 2\int_0^1 g_t$, we can easily find $g_t$ and $h_t$ with the supposed form as follows:

$$g_t(x):= \frac{1}{1-t}V(x-t),\quad h_t(x):= \frac{1}{1-t}V(x-2t+1).\tag{1}$$ It is easy to verify that $g_t$ and $h_t$ satisfy all the requirements in the statement for $f_t$.

When $f$ is an increasing $C^1$ function with $f(0)=0$, it can be expressed as $$f(x)=\int_0^1 f_t(x)f'(t)dt.$$ As a result, $g$ and $h$ can be constructed by expressing them in a similar way in terms of $g_t$ and $h_t$ respectively:

$$g(x):=\int_0^1 g_t(x)f'(t)dt,\quad h(x):=\int_0^1 h_t(x)f'(t)dt.\tag{2}$$ Due to $(2)$ and the fact $f'\ge 0$, it is easy to check that all the requirements for $g$ and $h$ are inherited from that for $g_t$ and $h_t$.

When $f$ is only increasing, the correct formulas of $g$ and $h$ can be guessed from $(1)$ and $(2)$ using integration by parts. Simple calculation shows that when $x\in[0,1)$, $$g(x)=(1-x)\int_0^x\frac{f(t)}{(1-t)^2}dt, \quad h(x)=(1-x)\int_0^{\frac{1+x}{2}} \frac{f(t)}{(1-t)^2}dt.\tag{3}$$

It is easy to check that $$g(1):=\lim_{x\to 1^-}g(x)=\lim_{x\to 1^-}f(x),\quad h(1):=\lim_{x\to 1^-}h(x)=2\lim_{x\to 1^-}f(x),\tag{4}$$ and $g$ and $h$ defined in $(3)$ and $(4)$ satisfy all the requirements.

• How did you get the idea of $\frac{x-2t+1}{1-t}$ ? – user119228 Mar 9 '14 at 21:53
• @Julien: I edited my answer by reorganizing the argument. I hope it looks clearer now. In the current form, your question asks how I get $h_t$ in $(1)$. First I assume $h_t(x)=b\cdot V(x-a)$ for some constants $a$, $b$. Then from $h_t\ge f_t$ we know $b(t-a)=h_t(t)\ge f_t(t)=1$. To get $\int_0^1h_t \le 2\int_0^1f_t=2(1-t)$, I use a stronger inequality $\frac{b(1-a)^2}{2}=\int_a^1 h_t\le 2(1-t)$ instead. The inequalities determine a unique pair of $(a,b)=(2t-1,\frac{1}{1-t})$. There is probably more geometrical explanation, but maybe it needs more words due to the limitation of my English. – 23rd Mar 10 '14 at 12:19
• Don't worry I am a language freak..Actually I have understood, it's very interesting how you get the idea. Thanks a lot. – user119228 Mar 10 '14 at 15:32
• @Julien: You are welcome! – 23rd Mar 10 '14 at 16:26