I am reading Evan's PDE book, and I need some help understanding the following result
[Theorem 1 on pg47 of the book]
Let $g$ be continuous and essentially bounded function on $\mathbb{R}^n$ and let $K$ be the heat kernel. Then, the function $u$ which is a convolution of $g$ and $K$ is $C^{\infty}$.
The proof of this theorem goes as:
Since $K$ is infinitely differentiable, with uniformly bounded derivatives of all orders, on $[\delta, \infty)$ for each $\delta > 0$, we see that $u$ is $C^{\infty}$
I am not really understanding this proof.
(1) Am I correct that the uniform boundedness of derivatives of all orders means that:
There exists a constant $M$ such that for every non-negative integer $\alpha$ and multi-index $\beta$, $|\frac{\partial^\alpha}{\partial t^{\alpha}} D^{\beta} K(x,t)| \leq M$ for every $x$ and $t$ in $\mathbb{R}^n \times [\delta, \infty)$?
If so, how do I know that the derivatives are uniformly bounded? I know that each derivative is bounded since $t\geq \delta > 0 $, but how do I show the existence of the uniform constant $M$?
(2) Why does the uniform boundedness of all derivatives allow us to differentiate under the integral sign? If I let $\Delta f_h$ to denote the difference quotient for the corresponding derivative $Df$, then I can write
$$\int |\Delta f_h(x-y)| g(y) dy = \int |Df(x-y+c)| g(y) dy$$ for some $c$ between $x-y$ and $x-y+h$ by the mean value theorem, so I have as my dominating function $M |g(y)|$ where $M$ is the uniform bound constant, but since $g$ is only bounded and not necessarily integrable, so I cannot apply the dominated convergence theorem. What am I doing wrong?
It was never intuiviely clear to me when differentiation under the integral is allowed and when it is not allowed. For example,
(3) suppose that I have a function $f(x,y)$ in $\mathbb{R}^2$ and assume further that we know $\frac{\partial}{\partial x} f(x,y)$ exists and is integrable in $y$ over $\mathbb{R}$. Then, is it always the case
$\dfrac{d}{dx} \int_{\mathbb{R}} f(x,y)dy = \int_{\mathbb{R}} \frac{\partial}{\partial x} f(x,y) dy$?
Best Answer
Let me give a simple example to illustrate what is required: Consider
$$g(x) = \int_{-\infty}^\infty f(x,y) \, dy.$$
The question is when can we exchange the derivative and integral, so that
$$g'(x) = \int_{-\infty}^\infty f_x(x,y) \, dy.$$
Taking difference quotients we have
$$\frac{g(x+h) - g(x)}{h} = \int_{-\infty}^\infty \frac{f(x+h,y)-f(x,y)}{h} \, dy.$$
So the question is really when can we exchange the limit as $h \to 0$ with the integral. To use the dominated convergence theorem, we need a dominating function that is integrable, so we need that for some $\delta>0$ there exists an integrable function $g(y)$ such that
$$\left|\frac{f(x+h,y)-f(x,y)}{h}\right| \leq g(y) \ \ \text{ for all } |h|<\delta.$$
By the mean value theorem
$$f(x+h,y) - f(x,y) = hf_x(z,y)$$
for some $z$ between $x$ and $x+h$. So it is enough to assume that for some $\delta>0$ there exists an integrable function $g(y)$ such that
$$|f_x(z,y)| \leq g(y) \ \ \text{for all } z \text{ with } |z-x|\leq \delta.$$
This condition is often called uniform integrability of $f_x$, meaning that $f_x$ is integrable uniformly in its first argument.
All derivatives of the heat kernel satisfy the uniform integrability property as long as you restrict $t$ away from zero. This is what Evans means when he says "uniform boundedness".