Integral and inequality

Question

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?

Answer 1

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 $\lambda=0$.

Answer 2

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} $