# Seems to be the optimal coefficient, $k$ integral inequality $\int_{0}^{1}f^2(x)dx\le \frac{423405}{246064}\left(\int_{0}^{1}f(x)dx\right)^2$

if $$f\in C^4[0,1]$$,such $$f(x)\ge 0,f'(0)\ge 0,f'(1)\le 0, f^{(4)}(x)\ge 0$$

show that $$\int_{0}^{1}f^2(x)dx\le\dfrac{423405}{246064}\left(\int_{0}^{1}f(x)dx\right)^2$$

My approach: it Seems to be looking for a polynomial $$P(x)$$,where $$\deg(P(x))=4$$,such $$\int_{0}^{1}P(x)dx=\dfrac{423405}{246064}，P(x)\ge 0,P'(0)\ge 0,P'(1)\le 0,$$We just have to prove the following inequality $$\int_{0}^{1}f^2(x)dx-\int_{0}^{1}P(x)dx\cdot \left(\int_{0}^{1}f(x)dx\right)^2\le 0$$ The next step would be to formulate the Integrand, or use Cauchy–Schwarz inequality, but that seems difficult

New add it: the $$\dfrac{423405}{246064}$$ is best

• How do you know $\dfrac{423405}{246064}$ is best? Mar 3, 2022 at 10:45

The right bound is

$$\int_0^1 f(x)^2\,dx ≤ \frac{52}{35}\left(\int_0^1 f(x)\,dx\right)^2.$$

which is tight; for example, $$f(x) = 3x^2 - 2x^3$$ attains equality.

To prove it, let

$$\begin{gather*} a = f(0) ≥ 0, \quad b = f'(0) ≥ 0, \quad c = f(1) ≥ 0,\quad d = -f'(1) ≥ 0, \\ p(x) = (1 - x)^2[a(1 + 2x) + bx] + x^2[c(3 - 2x) + d(1 - x)] \end{gather*}$$

be the cubic with $$p(0) = a$$, $$p'(0) = b$$, $$p(1) = c$$, $$p'(1) = -d$$, and let

$$r(x, y) = \begin{cases} x^2(1 - y)^2(3y - x - 2xy) \frac{f^{(4)}(y)}{6} & \text{if 0 ≤ x ≤ y ≤ 1}, \\ (1 - x)^2y^2(3x - y - 2xy) \frac{f^{(4)}(y)}{6} & \text{if 0 ≤ y ≤ x ≤ 1}. \end{cases}$$

For $$0 ≤ x, y ≤ 1$$, we have $$p(x) ≥ 0$$ and $$r(x, y) ≥ 0$$; we can check that

$$\int_0^1 r(x, y)\,dy = f(x) - p(x)$$

via repeated integration by parts. We also have

$$\begin{gather*} \int_0^1 p(x)\,dx = \frac{6a + b + 6c + d}{12}, \\ \begin{split} &\int_0^1 p(x)^2\,dx \\ &\quad = \frac{78a^2 + 22ab + 2b^2 + 54ac + 13bc + 78c^2 + 13ad + 3bd + 22cd + 2d^2}{210}, \end{split} \\ \begin{split} &\frac{52}{35}\left(\int_0^1 p(x)\,dx\right)^2 - \int_0^1 p(x)^2\,dx \\ &\quad = \frac{24ab + b^2 + 612ac + 78bc + 78ad + 8bd + 24cd + d^2}{1260} ≥ 0, \end{split} \\ \int_0^1 p(x)^2\,dx ≤ \frac{52}{35}\left(\int_0^1 p(x)\,dx\right)^2. \end{gather*}$$

Furthermore,

$$\int_0^1 r(x, y)\,dx = y^2(1 - y)^2 \frac{f^{(4)}(y)}{24},$$

and if $$0 ≤ y ≤ z ≤ 1$$,

$$\begin{split}\int_0^1 r(x, y)r(x, z)\,dx &= y^2(1 - z)^2(-y^5 + 7y^4z - 2y^5z + 12z^2 - 22yz^2 + 2z^3 \\ &\qquad + 8yz^3 - 8z^4 + 3yz^4 + 3z^5 - 2yz^5)\frac{f^{(4)}(y)f^{(4)}(z)}{5040} \\ &≤ 13y^2(1 - y)^2z^2(1 - z)^2\frac{f^{(4)}(y)f^{(4)}(z)}{5040} \\ &= \frac{52}{35}\int_0^1 r(x, y)\,dx \int_0^1 r(x, z)\,dx, \end{split}$$

and similarly if $$0 ≤ z ≤ y ≤ 1$$.

Therefore,

$$\begin{split} \int_0^1 f(x)^2\,dx &= \int_0^1 \int_0^1 \int_0^1 (p(x) + r(x, y))(p(x) + r(x, z))\,dx\,dy\,dz \\ &= \int_0^1 p(x)^2\,dx + 2\int_0^1\int_0^1 p(x)r(x, y)\,dx\,dy \\ &\qquad + \int_0^1\int_0^1\int_0^1 r(x, y)r(x, z)\,dx\,dy\,dz \\ &≤ \frac{52}{35}\left(\int_0^1 p(x)\,dx\right)^2 + 2⋅\frac{52}{35}\int_0^1\int_0^1 p(x)r(x, y)\,dx\,dy \\ &\qquad + \frac{52}{35}\left(\int_0^1\int_0^1 r(x, y)\,dx\,dy\right)^2 \\ &= \frac{52}{35}\left(\int_0^1\int_0^1 p(x) + r(x, y)\,dx\,dy\right)^2 \\ &= \frac{52}{35}\left(\int_0^1 f(x)\,dx\right)^2. \end{split}$$

• Nice! How did you arrive at the formula for $r(x,y)$ ? By trial and error ? Mar 4, 2022 at 14:56
• @EwanDelanoy I engineered $r(x, y)$ from the constraints $r(x, y) = \frac{∂}{∂x} r(x, y) = 0$ at $x = 0, 1$, and $\frac{∂^4}{∂^4x} r(x, y) = δ(y - x) f^{(4)}(y)$. Integrating this over $y$ yields $\int_0^1 \frac{∂^4}{∂^4x} r(x, y)\,dy = f^{(4)}(x)$, from we should be able to infer $\int_0^1 r(x, y)\,dy = f(x) - p(x)$, but I’m not sure if it’s easy to formalize an argument using boundary value problems and Dirac deltas. Mar 4, 2022 at 19:08