Can we prove that $\displaystyle 12 \int_{0}^{1}f^{2}(x)dx\le\int_{0}^{1}(1-x^4)(f''(x)) ^2dx$ if $f''$ is only Riemann integrable? Let $f:[0, 1]\to \mathbb{R}$ be a two times differentiable function such that $f''$ is Riemann integrable, $f(0)=0$ and $f'(0)=0$. Is it true under these assumptions that $$12 \int_{0}^{1}f^{2}(x)dx\le\int_{0}^{1}(1-x^4)(f''(x)) ^2dx?$$
The source of my problem is this, where I also proved this inequality if $f$ is a $C^2$ class function, i.e. if we know that $f''$ is not only Riemann integrable, but actually continuous. For the sake of completeness, I will add my short proof here.
By Taylor's Theorem with Integral Remainder we know that $$f(x)=\int_0^x (x-t)f''(t)dt$$
for any $x\in [0,1]$.
Thus, $$\int_0^1 f^2(x)dx=\int_0^1 \left(\int_0^x (x-t)f''(t) dt\right)^2 dx \stackrel{\text{C-S}}{\le}\int_0^1 \frac{x^3}{3}\left(\int_0^x (f''(t))^2 dt\right)dx=\frac{1}{12}\int_0^1(1-x^4)(f''(x))^2dx$$ and we are done.
The OP from AoPS required $f''$ to be only Riemann integrable and I wonder if this inequality still holds in that case or if he actually intended $f''$ to be continuous.
So far I've only come up with the idea that maybe I can somehow approximate a two times differentiable function with a Riemann integrable second derivative by a sequence of $C^2$ class functions. I don't know if this is possible (I haven't taken an advanced course in real analysis yet, so please bear with me if there is some well known density argument that I am not aware of), but this is all I could come up with.
 A: The Riemann integrability of $f''$ is sufficient to ensure that Taylor's theorem with the integral remainder holds. Using integration by parts, you have
$$
\int_0^x (x-t)f''(t) \, dt = \left[ (x-t)f'(t) \right]_{t=0}^{t=x} + \int_0^x f'(t) \, dt =- x f'(0) + f(x) - f(0)
$$
or
$$
 f(x) = f(0)  + xf(0) + \int_0^x (x-t)f''(t) \, dt \, .
$$
More generally, the integral form of the remainder
$$
 R_{k}(x)=\int _{a}^{x}{\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k}\,dt.
$$
is valid if $f^{(k)}$ is absolutely continuous on a closed interval $[a, b]$.
In order to show that
$$ \tag{*}
\int_0^1 \frac{x^3}{3}\left(\int_0^x (f''(t))^2 dt\right)dx=\frac{1}{12}\int_0^1(1-x^4)(f''(x))^2dx
$$
we can take a short detour via the Lebesgue integral. A Riemann integrable function on a compact interval is continuous almost everywhere, it is also Lebesgue integrable, and the Riemann integral coincides with the Lebesgue integral (see for example General condition that Riemann and Lebesgue integrals are the same).
Therefore,
$$
 g(x) = \frac{x^4-1}{12} \int_0^x f''(t)^2 \,dt
$$
is differentiable almost everywhere on $[0, 1]$, with
$$
g'(x) = \frac{x^3}3 \int_0^x f''(t)^2 \,dt - \frac{1-x^4}{12} f''(x)^2 \, .
$$
The right-hand side is Riemann- (and therefore Lebesgue-)integrable, so that the fundamental theorem of Lebesgue integral calculus can be applied, and it follows that
$$
0 = g(1) - g(0) = \int_0^1 g'(x) \, dx \\
= \int_0^1 \frac{x^3}{3}\left(\int_0^x (f''(t))^2 dt\right)dx -\frac{1}{12}\int_0^1(1-x^4)(f''(x))^2dx \, .
$$
Finally, all integrals on the right exist as Riemann integrals, so that $(*)$ holds for the Riemann integral as well.
A: You can also see that
$$
\int_0^1 {x^3\over 3}\left(\int_0^x (f''(t))^2 dt\right) dx={1\over 12}\int_0^1 (1-t^4)(f''(t))^2 dt
$$
by Fubini's theorem (interchange of the order of integration):
$$
\eqalign{
\int_0^1 {x^3\over 3}\left(\int_0^x (f''(t))^2 dt\right) dx
&= \int_0^1 \left(\int_t^1 {x^3\over 3} dx\right)(f''(t))^2 dt\cr
&=\int_0^1 {1-t^4\over 12}(f''(t))^2 dt.\cr
}
$$
For justification of the application of Fubini you can use the observation of Martin R that $f''$ is Lebesgue measurable and bounded, being Riemann integrable by hypothesis.
