In order to prove some results regarding Kolmogorov-Landau inequality for L2 norms, I am looking at the following integral differential equation:

$$\begin{equation*} \left(\int_{0}^{+\infty} f''^2\right)\left(\int_0^{+\infty} f^2\right) = \dfrac{1}{4} \int_{0}^{+\infty} f'^2 \qquad \text{(E)} \end{equation*}$$

With $f, f''$ square-integrable on $\mathbb{R}_{+}$, which implies square-integrability of $f'$ on $\mathbb{R}_{+}$.

The only approach I found to attempt to solve this differential equation:

  • express it as a limit of integrals, denote their antiderivatives and proceed to derivation to obtain a less "integral differential equation", which will fail because of the product.

I do not know a lot about these kinds of differential equations, any pointers on how to proceed?


Consider an arbitrary function $g$ square-integrable on $\mathbb{R}_{+}$

One can compute the constants $c_1,c_2,c_3 $ : $\quad\begin{cases} c_1=\int_{0}^{+\infty} g^2\\ c_2=\int_{0}^{+\infty} g'^2\\ c_3=\int_{0}^{+\infty} g''^2 \end{cases}$

Let $f=\sqrt{\frac{c_2}{4c_1c_3}}\:g$

$\quad\begin{cases} \int_{0}^{+\infty} f^2=\left(\sqrt{\frac{c_2}{4c_1c_3}}\right)^2\int_{0}^{+\infty} g^2=\left(\sqrt{\frac{c_2}{4c_1c_3}}\right)^2c_1=\frac{c_2}{4c_3}\\ \int_{0}^{+\infty} f'^2=\left(\sqrt{\frac{c_2}{4c_1c_3}}\right)^2\int_{0}^{+\infty} g'^2=\left(\sqrt{\frac{c_2}{4c_1c_3}}\right)^2c_2=\frac{c_2^2}{4c_1c_3}\\ \int_{0}^{+\infty} f''^2=\left(\sqrt{\frac{c_2}{4c_1c_3}}\right)^2\int_{0}^{+\infty} g''^2=\left(\sqrt{\frac{c_2}{4c_1c_3}}\right)^2c_3=\frac{c_2}{4c_1}\\ \end{cases}$

$$\left(\int_{0}^{+\infty} f''^2\right)\left(\int_0^{+\infty} f^2\right) =\frac{c_2}{4c_1}\frac{c_2}{4c_3}=\frac{c_2^2}{16c_1c_3}=\dfrac{1}{4} \int_{0}^{+\infty} f'^2$$

Thus, the general solution of the integral equation (E) is : $$f=\sqrt{\frac{c_2}{4c_1c_3}}\:g$$ where $g$ is any function square-integrable on $\mathbb{R}_{+}$ with the constants $c_1,c_2,c_3$ defined above.

  • $\begingroup$ Awesome, why did you consider $f = \sqrt{\dfrac{c_2}{4c_1c_3}} g$, is there any reason that motivated you to look at this particular form? $\endgroup$ – Raito Dec 29 '17 at 16:57
  • 1
    $\begingroup$ Obviously, the initial equation is a relationship between three constants. That is why I consider the general form $f=C g$ with arbitrary function $g$ and the coefficient $C$ is determined to satisfy the equality. $\endgroup$ – JJacquelin Dec 29 '17 at 17:15

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