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 $\dfrac{d}{dx} \int_{\mathbb{R}} f(x,y)dy = \int_{\mathbb{R}} \frac{\partial}{\partial x} f(x,y) dy$?

  • $\begingroup$ The constant in (1) can be different for every derivative. You should use the dominated convergence theorem to exchange limit as $h\to 0$ with integration to show derivative exists. What condition would you need to find a dominating function? $\endgroup$ – Jeff Feb 4 '17 at 16:29
  • $\begingroup$ @Jeff If the constant in (1) is different, in what sense the derivatieves are bounded "uniform"? $\endgroup$ – nan Feb 4 '17 at 19:24
  • $\begingroup$ Uniform on $[\delta,\infty)$. $\endgroup$ – Jeff Feb 4 '17 at 19:49
  • $\begingroup$ All the derivatives are uniformly integrable on $[\delta,\infty)$, and this is what you should use. $\endgroup$ – Jeff Feb 4 '17 at 19:51
  • $\begingroup$ @Jeff I am the OP still trying to understand what is going on. Why do I need uniform integrability of the derivatives? $\endgroup$ – yumiko Feb 6 '17 at 16:39

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".

  • $\begingroup$ Thanks so much for the explanation! Am I right then that the constant $M$ in my original question (1) which can be different for every derivative is not at all helpful? It would be useful if the measure of the set that I am integrating over is finite, would'n it? So, when Evan says "uniform bounded", he means uniformly+locally(in $x$) by an integrable function "$g(y)$"? $\endgroup$ – yumiko Feb 6 '17 at 17:07
  • 1
    $\begingroup$ Yes, uniform boundedness by a constant is only useful on finite measure sets (since you can use constant function in dominated convergence theorem). When Evans says "uniform boundedness", he means uniformly bounded by an integrable function (or uniformly integrable). $\endgroup$ – Jeff Feb 6 '17 at 18:44

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