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 ?

Thanks in advance,
 A: 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. 
