How to prove that there exists $g(x)$ such $\int_{0}^{1}g(x)dx\ge\frac{1}{2}\int_{0}^{1}f(x)dx$ let $f(x)\ge 0,x\in [0,1]$, and is increasing in $[0,1]$

show that: There exists $g(x)\ge 0,x\in [0,1]$,and  $g''(x)>0$, such
  $g(x)\le f(x)$, and such
  $$\int_{0}^{1}g(x)dx\ge\dfrac{1}{2}\int_{0}^{1}f(x)dx$$

I fell this maybe exists, see my following figure, But I can't How use methods prove it. and I think this is good problem, so I hope see someone can prove it, Thank you

 A: Firstly, note that given $f:[0,1]\to\Bbb R$, increasing, we can always define $f_l$ as follows: 
$$f_l(0):=f(0),\quad f_l(x):=\lim_{t\to x^-}f(t),\ \forall x\in(0,1].$$ 
Then $0\le f_l\le f$, $f_l$ is increasing and left continuous and $\int_0^1 f_l(x)dx=\int_0^1 f(x)dx$. Therefore, without loss of generality, in below we will assume that 
$$f:[0,1]\to[0,+\infty)\text{ is not constant, increasing and left continuous.} $$
Define 
$$\tilde{\Lambda}_f:=\{g:[0,1]\to[0,+\infty)\mid g\le f,\ g \text{ is convex} \},\quad \Lambda_f:=\{g\in \tilde{\Lambda}_f\mid g \text{ is increasing} \}$$
and define
$$\tilde{g}_f:=\sup_{g\in \tilde{\Lambda}_f}g\,,\quad
g_f:=\sup_{g\in \Lambda_f}g\,.$$
By definition, $\tilde{\Lambda}_f\supset\Lambda_f\ne \emptyset$, $\tilde{g}_f\in \tilde{\Lambda}_f$ and $g_f\in \Lambda_f$.

Proposition: If $f=c\cdot \chi_{(a,1]}$ for some $c>0$ and $a\in[0,1)$, then 
  $$\tilde{g}_f(x)=c\cdot\frac{x-a}{1-a}\cdot\chi_{(a,1]}(x)\quad\text{and}\quad\int_0^1 \tilde{g}_f(x)dx=\frac{1}{2} \int_0^1 f(x)dx;$$ otherwise, there
  exists $g\in\Lambda_f$, such that $g$ is smooth(so $g''\ge 0$) and
  $$\int_0^1 g(x)dx>\frac{1}{2} \int_0^1 f(x)dx.$$

Proof: If  $f$ is of the form $c\cdot \chi_{(a,1]}$, the conclusion is easy to check, so let us assume that $f$ does not have this form, and we claim that
$$\int_0^1 g_f(x)dx>\frac{1}{2}\int_0^1 f(x)dx.\tag{1}$$
Once $(1)$ is proved, we can use a convolution trick to complete the proof as follows.  Since $g_f$ is increasing and convex on $[0,1]$, we may extend it to a function $g_f:(-\infty,1]\to [0,+\infty)$ by defining $g_f(x)\equiv g_f(0)$ for $x\le 0$, which is still increasing and convex. Let $\rho:\Bbb R\to[0,+\infty)$ be some fixed smooth function supported on $[0,1]$ with $\int_{\Bbb R}\rho(x)dx=1$. For every $\delta>0$, define
$$g_\delta(x)=\int_0^1g_f(x-\delta t)\rho(t)dt=\frac{1}{\delta}\int_{-\infty}^1 g_f(s)\rho(\frac{x-s}{\delta})ds,\ \forall x\in[0,1].$$ 
By definiton, $g_\delta$ is  smooth, increaing, convex and $0\le g_\delta\le f$, i.e. $g_\delta\in \Lambda_f$. Moreover,
$$g_\delta(x)\ge g_f(x-\delta)\Longrightarrow \int_0^1g_\delta(x)dx\ge \int_{-\delta}^{1-\delta}g_f(x)dx.$$
Then by choosing $\delta$ small, the conclusion follows from $(1)$. Therefore, to complete the proof, it suffice to prove $(1)$.
Define
$$G_f:=\{x\in(0,1): f(x)>g_f(x)\}.$$
From the facts that $g_f\le f$, $g_f$ is continuous, and $f$ is increasing and left continous, we know that $G_f$ is open. If $G_f$ is empty, we are done; otherwise, let $(a,b)$ be a connected component of $G_f$, i.e. $g_f(a)=f(a)<f(b)=g_f(b)$ and $g_f(x)<f(x)$ on $\in(a,b)$, and we claim that $g_f$ is a linear function on $[a,b]$, i.e. 
$$g_f(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a),\ \forall x\in[a,b].\tag{2}$$ 
To show this, it suffices to show that for every $x_0\in(a,b)$, $f$ is linear in some open neighborhood of $x_0$. Since $g_f(x_0)<f(x_0)$, there exist $a<x_1<x_0<x_2<b$, such that 
$$f(t)>L(t):=g(x_1)+\frac{g(x_2)-g(x_1)}{x_2-x_1}(t-x_1),\ \forall t\in [x_1,x_2].$$
Note that if we define $g=g_f$ on $[0,x_1]\cup [x_2,1]$ and $g=L$ on $[x_1,x_2]$, then $g\in \Lambda_f$ and $g\ge g_f$, so by definition, $g=g_f$, which proves $(2)$. It follows that 
$$\int_a^bg_f(x)dx=\frac{f(a)+f(b)}{2}(b-a)\ge \frac{f(b)}{2}(b-a)\ge \frac{1}{2}\int_a^bf(x)dx,$$
and the equality holds if and only if $f(a)=0$ and $f(x)\equiv f(b)$ for $x\in(a,b]$. As a result,
$$\int_0^1g_f(x)dx= \int_{G_f}g_f(x)dx +\int_{[0,1]\setminus G_f}f(x)dx\ge\frac{1}{2}\int_0^1f(x)dx,$$
and when the equality holds, we must have (i) $\int_{G_f}g_f(x)dx=\frac{1}{2}\int_{G_f}f(x)dx$ and (ii) $\int_{[0,1]\setminus G_f}f(x)dx=0$. (i) and (ii) imply that $G_f=(a,1)$ for some $a\in[0,1)$ and $f(a)=0$, i.e. $f=f(1)\cdot \chi_{(a,1]}$, which completes the proof. $\quad\square$.

Remark: In an earlier version of this answer it was proved that there exists an explicit way to construct some $g\in \Lambda_f$ with $\int_0^1g(x)dx=\frac{1}{2}\int_0^1f(x)dx$ as follows.
$$g(x):=(1-x)\int_0^x \frac{f(t)}{(1-t)^2}dt, \  x\in[0,1),\quad g(1):=\lim_{x\to 1^-}g(x)=\lim_{x\to 1^-}f(x).$$
