# Integral and inequality

Let $$p(u,x):=(4 \pi u)^{-1/2}e^{-\frac{x^2}{4u}},u>0,x \in \mathbb{R}.$$

Let $$\phi \in C_c^{\infty}(\mathbb{R}),\text{supp}(\phi) \subset B(0,1),||\phi||_{\infty} \leq 1.$$

Prove that for all $$U>0,\beta>1/2,$$ there exist $$\epsilon>0,C>0$$ such that for all $$u\in [0,U],\lambda \in \left]0,1\right],$$ $$\int_0^{u} \int_{\mathbb{R}} \left(\int_{\mathbb{R}} \phi^\lambda(y_1)p(r,y_1-y_2) dy_1 \right)^2 dy_2 dr\leq Cu^\varepsilon \lambda^{1-2\beta},$$ where $$\phi^\lambda(y) = \lambda^{-1} \phi(\lambda^{-1}y).$$

I tried, using a change of variable, replacing $$\phi^{\lambda}$$ with $$\phi.$$ also $$\lambda(B(0,1))<\infty$$ might be useful.

How can we prove this inequality?

Define \begin{aligned} I_1&=\int_{-\infty}^\infty I_2^2\text{ d}y_2\\ I_2&=\int_{-\infty}^\infty\frac1\lambda\phi\left(\frac{y_1}{\lambda}\right)\cdot\frac1{\sqrt{4\pi r}}e^{-(y_1-y_2)^2/4r}\text{ d}y_1\\ \end{aligned} then the result we want to show is that $$I=\int_0^{u}I_1\text{ d}r\leq Cu^\epsilon\lambda^{1/2-\beta}$$ First, we'll simplify $$I_2$$. Since $$\phi(x)$$ is nonzero only for $$-1\leq x\leq 1$$, $$\phi(\frac{y_1}{\lambda})$$ is nonzero only for $$-1\leq \frac{y_1}{\lambda}\leq 1$$, or $$-\lambda\leq y_1\leq\lambda$$. Therefore, $$I_2=\frac1{\lambda\sqrt{4\pi r}}\int_{-\lambda}^{\lambda}\phi\left(\frac{y_1}{\lambda}\right)e^{-(y_1-y_2)^2/4r}\text{ d}y_1$$ Next, expand out $$I_1$$, and do the $$y_2$$ integral first: (the validity of interchanging the integrals follows from $$|\phi(x)|\leq1$$) \begin{aligned} I_1&=\frac1{4\pi r\lambda^2}\int_{-\infty}^\infty\left(\int_{-\lambda}^{\lambda}\phi\left(\frac{y_1}{\lambda}\right)e^{-(y_1-y_2)^2/4r}\text{ d}y_1\right)\left(\int_{-\lambda}^{\lambda}\phi\left(\frac{y_1'}{\lambda}\right)e^{-(y_1'-y_2)^2/4r}\text{ d}y_1'\right)\text{ d}y_2\\ &=\frac1{4\pi r\lambda^2}\int_{-\lambda}^{\lambda}\int_{-\lambda}^{\lambda}\phi\left(\frac{y_1}{\lambda}\right)\phi\left(\frac{y_1'}{\lambda}\right)\int_{-\infty}^\infty\exp\left(-\frac{(y_2-y_1)^2+(y_2-y_1')^2}{4r}\right)\text{ d}y_2\text{ d}y_1\text{ d}y_1' \end{aligned} The innermost integral may be evaluated as \begin{aligned} \int_{-\infty}^\infty\exp\left(-\frac{(y_2-y_1)^2+(y_2-y_1')^2}{4r}\right)\text{ d}y_2&=e^{-(y_1-y_1')^2/8r}\int_{-\infty}^\infty\exp\left(-\frac1{2r}\left(y_2-\frac{y_1+y_1'}2\right)^2\right)\text{ d}y_2\\ &=e^{-(y_1-y_1')^2/8r}\sqrt{2\pi r} \end{aligned} Substituting this back into $$I_1$$ gives \begin{aligned} I_1&=\frac1{2\sqrt{2\pi r}\lambda^2}\int_{-\lambda}^{\lambda}\int_{-\lambda}^{\lambda}\phi\left(\frac{y_1}{\lambda}\right)\phi\left(\frac{y_1'}{\lambda}\right)e^{-(y_1-y_1')^2/8r}\text{ d}y_1\text{ d}y_1'\\ &\leq\frac1{2\sqrt{2\pi r}\lambda^2}\int_{-\lambda}^{\lambda}\int_{-\lambda}^{\lambda}e^{-(y_1-y_1')^2/8r}\text{ d}y_1\text{ d}y_1'\\ &\leq\frac1{2\sqrt{2\pi r}\lambda^2}\int_{-\lambda}^{\lambda}\int_{-\lambda}^{\lambda}\text{ d}y_1\text{ d}y_1'\\ &=\frac{4\lambda^2}{2\sqrt{2\pi r}\lambda^2}\\ &=\sqrt{\frac2\pi}r^{-1/2} \end{aligned} where we have used the bounds $$|\phi|\leq1$$ and $$e^{-x^2}\le1$$. Note that equality holds only when $$\lambda=0$$. Integrating $$I_1$$, we arrive at \begin{aligned} I&=\int_0^{u}I_1\text{ d}r\\ &\le\sqrt{\frac2\pi}\int_0^{u}r^{-1/2}\text{ d}r\\&=2\sqrt{\frac2\pi}u^{1/2} \end{aligned} This shows that as long as you pick $$C\ge2\sqrt{2/\pi},\epsilon\ge1/2$$, we will always have $$I\le2\sqrt{\frac2\pi}u^{1/2}\le Cu^{1/2}\le Cu^{1/2}\lambda^{1-2\beta}$$ since $$\lambda^{1-2\beta}\ge1$$ for all $$0\le\lambda\le1,\beta>1/2$$. By the way, if you still need a tighter bound, you can actually compute the double integral in the first inequality for $$I_1$$ explicitly in terms of error functions: \begin{aligned} I_1&=\frac1{2\sqrt{2\pi r}\lambda^2}\int_{-\lambda}^{\lambda}\int_{-\lambda}^{\lambda}\phi\left(\frac{y_1}{\lambda}\right)\phi\left(\frac{y_1'}{\lambda}\right)e^{-(y_1-y_1')^2/8r}\text{ d}y_1\text{ d}y_1'\\ &\leq\frac1{2\sqrt{2\pi r}\lambda^2}\int_{-\lambda}^{\lambda}\int_{-\lambda}^{\lambda}e^{-(y_1-y_1')^2/8r}\text{ d}y_1\text{ d}y_1'\\ &=\frac1{2\sqrt{2\pi r}\lambda^2}\left(4\sqrt{2\pi r}\lambda\text{ erf}\left(\frac{\lambda}{\sqrt{2r}}\right)+8r\left(e^{-\lambda^2/2r}-1\right)\right)\\ &=\frac2\lambda\text{ erf}\left(\frac{\lambda}{\sqrt{2r}}\right)+\frac{2}{\lambda^2}\sqrt{\frac{2r}{\pi}}\left(e^{-\lambda^2/2r}-1\right) \end{aligned} Integrating with respect to $$r$$, we obtain \begin{aligned} I&=\int_0^{u}I_1\text{ d}r\\ &\le\int_0^{u}\frac2\lambda\text{ erf}\left(\frac{\lambda}{\sqrt{2r}}\right)+\frac{2}{\lambda^2}\sqrt{\frac{2r}{\pi}}\left(e^{-\lambda^2/2r}-1\right)\text{ d}r\\ &=\frac{2u}\lambda\text{ erf}\left(\frac{\lambda}{\sqrt{2u}}\right)-\frac23\lambda \left(1-\text{ erf}\left(\frac{\lambda}{\sqrt{2u}}\right)\right)+\frac{4}{3}\sqrt{\frac{2}\pi}\frac{u^{3/2}}{\lambda^{2}}\left(e^{-\lambda^2/2u}-1\right)+\frac23\sqrt{\frac{2u}\pi}e^{-\lambda^2/2u} \end{aligned} This is actually the tightest possible bound for $$I$$ in terms of just $$u,\lambda$$ since you can construct bump functions that become arbitrarily close to $$1$$ over $$(-1,1)$$. Therefore, you can always find $$\phi$$ such that $$I$$ becomes arbitrarily close to this bound. Also, you can verify that this expression equals $$2\sqrt{2/\pi}u^{1/2}$$ when $$\lambda=0$$.

• It's possible to generalize this to $\mathbb{R}^2,$ by considering $p(u,x):=(4 \pi u)^{-1}e^{-\frac{x^2+y^2}{4u}},u>0,x,y \in \mathbb{R},$ and we should obtain $\lambda^{-2\beta}$ on the RHS. Do you have ideas how to prove it? the problem is that $\int_0^u1/rdr$ is not integrable. Commented Feb 3, 2023 at 15:25
• Can you give more details about your generalisation? There are many ways to generalise to $\mathbb R^2$. For example, would $\phi(x)$ be generalised as $\phi(x,y)$ or $\phi(x)\phi(y)$? It would be better if you could post the precise question as an edit or as a new question. Commented Feb 4, 2023 at 2:19
• We are considering in $\mathbb{R}^2,p(u,x,y):=(4 \pi u)^{-1}e^{-\frac{x^2+y^2}{4u}},u>0,x ,y\in \mathbb{R}.$ $\phi \in C_c^{\infty}(\mathbb{R}^2),\text{supp}(\phi) \subset B(0,1),||\phi||_{\infty} \leq 1.$ In this case we have $\int_0^{u} \int_{\mathbb{R}^2} \left(\int_{\mathbb{R}^2} \phi^\lambda(y_1)p(r,y_1-y_2) dy_1 \right)^2 dy_2 dr\leq Cu^\varepsilon \lambda^{-2\beta},$ how can we prove this generalization? The problem is that $\int^u_01/rdr$ is not integrable is not integrable. Commented Feb 4, 2023 at 3:23
• Where $\phi^{\lambda}(x)=\lambda^{-2}\phi(x/\lambda),x\in\mathbb{R}^2.$ Commented Feb 4, 2023 at 5:59
• This would be solved if we could prove $\forall u\in [0,U],\lambda\in]0,1],\int_0^u\int_{[-1,1]^2}\int_{[-1,1]^2}\frac{1}{r}e^{-\lambda^2|x-y|^2/r}dxdydr\leq Cu^{\epsilon}\lambda^{-2\beta}, |.|$ the euclidean norm Commented Feb 4, 2023 at 20:20

Let $$U>0$$ and $$\beta\ge1$$ (see comment below 1st paragraph).

\begin{align} \left(\int_{\mathbb{R}} \phi^\lambda(y_1)p(r,y_1-y_2) dy_1\right)^2 & =\left(\int_{\mathbb{R}}\phi^\lambda(y_1)(4\pi r)^{-1/2}e^{-(y_1-y_2)^2/(4r)} dy_1\right)^2 \\ & = \left((4\pi r)^{-1/2} \left|\int_{\mathbb{R}}\phi^\lambda(y_1)e^{-(y_1-y_2)^2/(4r)} dy_1\right|\right)^2 \\ & \le \left((4\pi r)^{-1/2} \left[\int_{\mathbb{R}}\left|\phi^\lambda(y_1) \right|^2 dy_1\ \right]^{1/2}\left[\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right]^{1/2} \right)^2 \text{(Schwarz Inequality)} \\ & = \left((4\pi r)^{-1/2}\right)^2 \left(\left[\int_{\mathbb{R}}\left|\phi^\lambda(y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2\left(\left[\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right]^{1/2} \right)^2 \\ & = \left|(4\pi r)^{-1}\right|\left(\left[\int_{\mathbb{R}}\left|\phi^\lambda(y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2\left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right| \\ & = \left|(4\pi r)^{-1}\right|\left(\left[\int_{\mathbb{R}}\left|\phi^\lambda(y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right|\\ & = \left|(4\pi r)^{-1}\right|\left(\left[\int_{-\infty}^\infty\left|\phi^\lambda(y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right|\\ & = \left|(4\pi r)^{-1}\right|\left(\left[\lim_{b\rightarrow\infty}\int_{-b}^b\left|\phi^\lambda(y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right|\\ & = \left|(4\pi r)^{-1}\right|\left(\lim_{b\rightarrow\infty}\left[\int_{-b}^b\left|\phi^\lambda(y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right| \text{(square root limit rule)}\\ & = \left|(4\pi r)^{-1}\right|\left(\lim_{b\rightarrow\infty}\left[\int_{-b}^b\left|\lambda^{-1}\phi(\lambda^{-1}y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right| \\ & = \left|(4\pi r)^{-1}\right|\left(\lim_{b\rightarrow\infty}\left[\int_{-b}^b\left|\lambda^{-1}\right|^2\left|\phi(\lambda^{-1}y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right|\\ & = \left|(4\pi r)^{-1}\right| ||\lambda^{-1}|| \left(\lim_{b\rightarrow\infty}\left[\int_{-b}^b\left|\phi(\lambda^{-1}y_1) \right|^2 dy_1\ \right]^{1/2}\right)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right| \\ & = \left|(4\pi r)^{-1}\right| |\lambda^{-1}|(||\phi||_\infty)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right| \\ & \le \left|(4\pi r)^{-1}\right|\lambda^{-1}(1)^2 \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right|\\ & = (4\pi r)^{-1}\lambda^{-1} \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right| \text{(since r>0)}\\ & = (4\pi r)^{-1}\lambda^{1-2} \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right|\\ & \le (4\pi r)^{-1}\lambda^{1-2\beta} \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right|\\ \end{align} (holds for all $$\lambda\in ]0,1]$$ and all $$\beta \ge 1$$; if $$1/2<\beta<1$$, then the inequality does not hold unless $$\lambda=1$$)

\begin{align} \int_0^u\int_{\mathbb{R}}\left(\int_{\mathbb{R}} \phi^\lambda(y_1)p(r,y_1-y_2) dy_1\right)^2 dy_2 \ dr & =\int_0^u\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\phi^\lambda(y_1)(4\pi r)^{-1/2}e^{-(y_1-y_2)^2/(4r)} dy_1\right)^2 dy_2 \ dr\\ &\le \int_0^u\int_{\mathbb{R}} (4\pi r)^{-1} \lambda^{1-2\beta} \left|\int_{\mathbb{R}}\left|e^{-(y_1-y_2)^2/(4r)}\right|^2dy_1 \right| dy_2 \ dr\\ &= \lambda^{1-2\beta} \int_0^u Cu^{\varepsilon-1} \ dr\\ &=Cu^{\varepsilon-1} \lambda^{1-2\beta} \int_0^u \ dr\\ &=Cu^{\varepsilon-1} \lambda^{1-2\beta} u\\ &=Cu^{\varepsilon} \lambda^{1-2\beta}\\ \end{align}