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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.

Alexdanut
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2 Answers2

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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.

Martin R
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    Yes, I know that Taylor's Theorem works even if $f''$ is only Riemann integrable. The problem is here: $\displaystyle \int_0^1 \frac{x^3}{3}\left(\int_0^x (f''(t))^2 dt\right)dx=\frac{1}{12}\int_0^1(1-x^4)(f''(x))^2dx$. I use integration by parts for this equality and if $f''$ is only Riemann integrable then I cannot say that $\displaystyle \left(\int_0^x (f''(t))^2 dt\right)'=(f''(x))^2$ (well, I actually only need $(f'')^2$ to be continuous, but whatever). – Alexdanut Aug 18 '21 at 14:01
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    @Alexdanut: The identity $\Big(\int^x_0 (f'')^2\Big)'=(f'')^2$ occurs almost everywhere (the exceptional points are points of discontinuity which form a set of measure zero by the assumption of $f''$) – Mittens Aug 18 '21 at 14:17
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    @Alexdanut: I have updated the answer. – Martin R Aug 18 '21 at 14:27
  • @MartinR thank you, so I think that what you are saying is that if I have two functions $u, v$ defined on $[0, 1]$ such that $u$ is differentiable and its derivative is Riemann integrable and $v$ is differentiable almost everywhere and its derivative is Riemann integrable, then I may still apply integration by parts, i.e. $\displaystyle \int_0^1 u'v=uv\bigg |_0^1 -\int_0^1 uv'$, right? Can you explain to me why this works, please? I don't know too much about the Lebesgue integral, but in this context is the integral $\displaystyle \int_0^1 uv'$ the usual Riemann integral? – Alexdanut Aug 18 '21 at 14:37
  • @Alexdanut: If $f$ in Riemann integrable over $[a,b]$ and $\alpha$ has derivative which is Riemann integrable over $[a,b]$, the the Riemann Sieltjes integral $\int^b_a f(x),d\alpha(x)$ exists and $\int^b_a f(x),d\alpha(x)=\int^b_a f(x)\alpha'(x),dx$ (Theorem 7.35 in Apostol's Math. Analysis). The rest is integration by parts for Riemann-Stieltjes integrals (Theorem 7.6, op.cit) – Mittens Aug 18 '21 at 14:42
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    @Alexdanut: I was thinking in terms of the Lebesgue integral: $g(x) = \frac{x^4-1}{12} \int_0^x f''(t)^2 ,dt$ is differentiable almost everywhere and the derivative is Riemann (and therefore Lebesgue) integrable, compare https://en.wikipedia.org/wiki/Absolute_continuity#Equivalent_definitions. It follows that $0 = g(1) - g(0) = \int_0^1 g'(x) dx$, which is the desired formula. What Oliver said might be an easier way to get the result. – Martin R Aug 18 '21 at 14:46
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    @Alexdanut: I have added more details. (Again, using the Riemann-Stieltjes integral as suggested by Oliver may be a simpler approach.) – Martin R Aug 18 '21 at 15:05
  • @MartinR thank you very much, now it all makes sense even for someone like me who is not so familiar with the Lebesgue integral! – Alexdanut Aug 18 '21 at 15:07
  • @OliverDiaz Thank you, I will certainly look into your approach using the Riemann-Stieltjes integral too, even though I have had little to none exposure to this type of integrals. – Alexdanut Aug 18 '21 at 15:10
  • It is very similar to The Riemann integral (where you use $\alpha(x)=x$) The theory works well for $\alpha$'s of bounded variation, which in essence means that $\alpha$ is the difference of two monotone nondecreasing functions. In your problem you have $\alpha(x)=f'(x)$ Since $f''$ is Riemann integrable, $f'$ will be of finite variation (over $[a,b]$) In fact, it will be absolutely continuous too. – Mittens Aug 18 '21 at 15:14
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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.

John Dawkins
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