TITLE: Some Scaling Estimate for Heat Kernel
QUESTION [5 upvotes]: NOTE. I have rewritten the question to summarize my current progress on this question.  The bounty is for completing what I have done so far, or by offering a more elegant solution probably based on some clever scaling/translation argument.  I have also slightly (I hope) clarified the wording in (b).
Let $$G(x,t)=\frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{|x|^{2}}{4t}}.$$ be the usual heat kernel.
(a) Given $\alpha>0$ find constants $\beta$ and $C$ so that
$$G(x+y,t)\leq CG(x,\beta t)$$
holds for every $x\in\mathbb{R}^{d}$, $t>0$ and $|y|\leq\alpha\sqrt{t}.$}
(b) Deduce that for $f\in L^{1}$, $\lambda>0$, and $u(x,t)=(G(t,\cdot)*f)(x)$, that
$$\mu\left(\left\{y:\text{s.t.}\;\exists t>0\;\text{whereby the estimate}|u(x,t)|\geq\lambda\;\text{holds whenever}\;x\in B(y,\alpha\sqrt{t})\right\}\right)\leq\frac{||f||_{L^{1}}}{\lambda}.$$

Let us for now assume $d=1$ so
$$G(x,t)=(4\pi t)^{-1/2}\exp\left(\frac{-x^{2}}{4t}\right).$$
After fixing $x\in\mathbb{R}$, $t>0$, $\alpha>0,$ and restricting $|y|\leq\alpha\sqrt{t},$ we are tasked to prove the estimate
$$(4\pi t)^{-1/2}\exp\left(\frac{-|x+y|^{2}}{4t}\right)\leq C(4\pi\beta t)^{-1/2}\exp\left(\frac{-|x|^{2}}{4\beta t}\right),$$
for constants $\beta=\beta(\alpha)>0$ and $C=C(\alpha)>0.$  Cancelling the factor $(4\pi t)^{-1/2}$ from both sides reduces the estimate to
$$\exp\left(\frac{-|x+y|^{2}}{4t}\right)\leq C\beta^{-1/2}\exp\left(\frac{-|x|^{2}}{4\beta t}\right).$$
Since we can multiply $C$ by $\beta^{1/2}$, we can absorb the factor $\beta^{-1/2}$ into $C$.  The estimate then further reduces to
$$\exp\left(\frac{-|x+y|^{2}}{4t}\right)\leq C\exp\left(\frac{-|x|^{2}}{4\beta t}\right).$$
Since $1/(4t)$ appears in both exponentials in a symmetric way, we can also ``cancel'' them to finally get
$$\exp\left(-|x+y|^{2}\right) \leq C\exp\left(-|x|^{2}/\beta\right).$$
Let us maximize the left hand side subject to the constraint $|y|\leq\alpha\sqrt{t}$, for clearly the validity of the resulting estimate will imply the present one.  The left hand side is maximized when $y=-x$, and this is valid if $|x|\leq\alpha\sqrt{t}$.  However, if $|x|>\alpha\sqrt{t}$, then $y=-\text{sgn}(x)\alpha\sqrt{t}$ is the best that can be done.   Thus we get the system of inequalities
$$\left\{\begin{array}{ll}
1\leq C\exp\left(-|x|^{2}/\beta\right)&\text{if}\;|x|\leq\alpha\sqrt{t},\\
\exp\left(-|x-\text{sgn}(x)\alpha\sqrt{t}|\right)\leq C\exp\left(-|x|^{2}/\beta\right)&\text{if}\;|x|>\alpha\sqrt{t}.\end{array}\right.$$
At this point I get the sense that I am going in the wrong direction, especially since we need to solve for $\beta$ and $C$ independently of $x$ and $t$.  Perhaps I should have kept the $1/(4t)$ factor in the exponentials.  Even when I did this separately, I still ended up with a similar set of inequalities with know obvious way to get rid $x$ and $t$ in them.
Any help extending this argument, or providing a more elegant alternative would be highly welcomed.
As for (b), the wording and all of the parameters involved still confuses me.  However, I can at least start by writing out $u(x,t)$ as
$$u(x,t)=\int_{-\infty}^{\infty}(4\pi t)^{-1/2}\exp\left(\frac{-|x-y|^{2}}{4t}\right)f(y)\;dy.$$
The appearance of the $|x-y|$ in the convolution already suggests (a) will be helpful.  Since $G(x,t)\in L^{1}$ for every $t>0$ and $f\in L^{1}$ by assumption, we see $u\in L^{1}$ for every $t>0$.  Thus $f$ satisfies the weak-type estimate (special case of Chebyshev inequality)
$$\mu\left(\{x:|u(x,t)|>\lambda\}\right)\leq\frac{||f||_{1}}{\lambda}.$$
The key then seems to be relating the above set with the one in (b).

REPLY [2 votes]: I posted this same kind of analysis for someone studying the Poisson kernel a couple of days ago. For vectors $x,y \in \mathbb{R}^{d}$,
$$
   |x|^{2}=|x+y-y|^{2}=|x+y|^{2}-2(x+y)\cdot y+|y|^{2} \le |x+y|^{2}+2|x+y||y|+|y|^{2}.
$$
Because $2ab \le a^{2}+b^{2}$ for any real numbers $a$, $b$, then
$$
   |x|^{2} \le |x+y|^{2}+|x+y|^{2}+|y|^{2}+|y|^{2}=2|x+y|^{2}+2|y|^{2}.
$$
Assuming $|y|\le \alpha \sqrt{t}$,
$$
      \frac{1}{2}|x|^{2} \le |x+y|^{2}+\alpha^{2}t
$$
$$
           -|x+y|^{2}\le -\frac{1}{2}|x|^{2}+\alpha^{2}t
$$
So $G(x,t)=(4\pi t)^{-d/2}e^{-\frac{|x|^{2}}{t}}$ satisfies
$$
      G(x+y,t) =\frac{1}{(4\pi t)^{d/2}}e^{-\frac{|x-y|^{2}}{t}} \le e^{\alpha^{2}}2^{d/2}\frac{1}{(4\pi (2t))^{d/2}}e^{-\frac{|x|^{2}}{2t}}=e^{\alpha^{2}}2^{d/2}G(x,2t).
$$
Just to be explicit, $\beta=2$ and $C=e^{\alpha^{2}}2^{d/2}$ are constants that will work.