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If $f:[a,b] to mathbbR, f(a)=0,f(b)=1$ is a convex increasing differentiable function on the interval $[a,b]$ . Prove that
$$int_a^bf^2(x),dxle frac23int_a^bf(x),dx$$

Since f is convex and increasing so $f''(x)ge 0 $ and $f'(x)ge 0$. Then I consider a function $g:[a,b]to mathbbR$, $g(x)=frac23int_a^xf(t),dt-int_a^xf^2(t),dt$. Now $f$ is differentiable implies $g$ is also but can't conclude $g'(x)ge 0$.

In Prove $int _0^infty f^2 dx leq cdots $ for $f$ convex the following theorem was shown:

If $F$ is convex and non-negative on $[0, infty)$ then $$
int _0^infty F^2(x) dx leq frac23cdot max_x in mathbb R^+ F(x) cdot int _0^infty F(x) dx , .$$

Our function $f$ is non-negative and convex on $[a, b]$ with $f(a) = 0$ and $f(b) = 1$. If we define $F$ on $[0, infty)$ as
$$
F(x) = begincases
f(b-x) & text for 0 le x le b-a \
0 & text for x > b-a
endcases
$$
then $F$ satisfies the hypotheses of the above theorem, and therefore
$$
int_a^bf^2(x),dx = int _0^infty F^2(x) dx leq frac23cdot max_x in mathbb R^+ F(x) cdot int _0^infty F(x) dx = frac23int_a^bf(x),dx , .
$$

Alternatively we can modify the proof of the above theorem for this case.
Define $varphi: [a, b] to Bbb R$ as
$$
varphi(x) = frac 23 f(x) int_a^x f(t) , dt - int_a^x f^2(t) , dt , .
$$
The goal is to show that $varphi$ is (weakly) increasing. Then the desired conclusion follows with
$$
0 = varphi(a) le varphi(b) = frac 23 int_a^b f(t) , dt - int_a^b f^2(t) , .
$$
Since $f$ is assumed to be differentiable, we have
$$
varphi'(x) = frac 23 f'(x) int_a^x f(t) , dt + frac 23 f^2(x) - f^2(x) \
= frac 23 f'(x) int_a^x f(t) , dt - frac 13 f^2(x) , .
$$
Now we distinguish two cases:

• If $f'(x) =0$ then $f'(t) =0$ for $a le t le x$, so that $f(x) = f(a) = 0$ and therefore $varphi'(x) = 0$.


• If $f'(x) >0$ then we estimate $f(t)$ from below by the tangent at $(x, f(x))$:
  $$
  int_a^x f(t) , dt ge int_x-f(x)/f'(x)^x bigl( f(x) + (t-x)f'(x) bigr) , dt = fracf^2(x)2f'(x)
  $$
  and therefore $varphi'(x) ge 0$.

So $varphi'(x) ge 0$ for all $x in [a, b]$, which means that $varphi$ is increasing on the interval, and we are done.

Remark 1: The proof becomes easier if we assume that $f$ is twice differentiable. Then
$$
varphi''(x) = frac 23 f''(x) int_a^x f(t) , dt ge 0
$$
so that $varphi'(x) ge varphi'(0) = 0$.

Remark 2: The proof works even without the assumption that $f$ is differentiable: As a convex function, $f$ has a right derivative
$$
f_+'(x) = lim_substackh to 0\ h > 0 fracf(x+h)-f(x)h
$$
everywhere in $[a, b)$, and we can replace $f'$ by $f_+'$ and $varphi'$ by $varphi_+'$ in the above argument.

Without any loss of generality, we shift and scale to set $a=0, b=1$. And now we consider the integrals, $ int_0^1f(x)dx $ and $ int_0^1f^2(x) dx $.

Convexity of $ f(x) $ assures that $f(x)leq x$. (1)

Now, we write the integrals as limits of Riemann sums.$ int_0^1f^2(x) dx = lim_h to 0, N to inftysum_r=0^N(f^2(rh) times h)$ $int_0^1f(x) dx = lim_h to 0, N to inftysum_r=0^N(f(rh) times h) $

$ f^2(rh) / f(rh) =f(rh) leq rh$ (from (1)). For this ratio to be maximum (that is when the ratio $ frac int_0^1f^2(x) dxint_0^1f(x) dx$ is maximum), $ f(rh)=rh $ for all $r$. $Rightarrow f(x)=x $. (2)

This means $ frac int_0^1f^2(x) dxint_0^1f(x) dx leq frac int_0^1x^2 dxint_0^1x dx = frac23$

Edit: The maximisation holds if a unique maximum function exists in every interval of (2). This holds if $f(x)$ is convex. Otherwise as correctly pointed out by Martin, this ratio can be more than $frac23$. For example, if $ f(x)= sin^2(fracpi x2)$, this ratio is 3/4.