# Question about a proof in Evans

On page 57. in Partial Differential Equation by Lawrence C. Evans, he prove the maximum principle for the Cauchy problem of the heat equation, i.e. (I quote)

Suppose $u\in C^2_1(\mathbb{R}^n\times (0,T])\cap C(\mathbb{R}^n\times [0,T])$ solves $u_t-\Delta u= 0$ in $\mathbb{R}^n\times (0,T)$ and $u=g$ on $\mathbb{R}^n\times \{t=0\}$. Moreover, u satisfies the growth estimate

$$u(x,t)\le Ae^{a|x|^2}$$

for $x\in\mathbb{R}^n,0\le t\le T$ for constants $A,a>0$. Then

$$\sup_{\mathbb{R}^n\times [0,T]}u = \sup_{\mathbb{R}^n}g$$

In the proof they define $v(x,t):=u(x,t)-\frac{\mu}{(T+\epsilon -t)^\frac{n}{2}}\exp{\frac{|x-y|^2}{4(T+\epsilon -t)}}$ The proof consists of several steps. First they show for $4aT<1$ that

1. $\max_{\overline{U_T}} v= \max_{\Gamma_T}v$, where $U_T:=B^0(y,r)\times (0,T]$ for fixed $r>0$.
2. If $x\in \mathbb{R}^n$ then $v(x,0)\le g(x)$

Now in equation $(29)$ they say: for $r$ selected sufficiently large, we have $v(x,t)\le A\exp{a(|y|+r)^2}-\mu (4(a+\gamma))^{\frac{n}{2}}\exp{(a+\gamma)r^2}\le \sup_{\mathbb{R}^n}g$. Why is this all less or equal the supremum of $g$, for large $r$?

And why can we conclude with all these facts, that $v(y,t)\le \sup_{\mathbb{R}^n} g$ for all $y\in \mathbb{R}^n$ and $0\le t\le T$?

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Hint: compare the coefficients of $r^2$ in the two exponential functions. –  Siminore Jun 18 '12 at 9:00
$$v(x,t)\le A\exp{(a(|y|+r)^2)}-\mu (4(a+\gamma))^{\frac{n}{2}}\exp{((a+\gamma)r^2)}=\exp{((a+\gamma)r^2)}[(-\mu (4(a+\gamma))^{\frac{n}{2}}+A\exp{(-\gamma r^2+2ar|y| + a|y|^2)]}$$
This converges to $-\infty$ as $r\to \infty$ and the conclusion follows.
We already know that the max is attained at the "boundary" $\Gamma_T$ and we have found a bound of $v$ by (the supremum) of $g$ on $\Gamma_T$.