Help with a DCT problem The question is

Let ${p_t}\left( x \right) = \frac{1}{{\sqrt {2\pi t} }}{e^{ - \frac{{{x^2}}}{{2t}}}},t > 0,x \in \Bbb{R}$. It is known that $\int_R {\frac{1}{{\sqrt {2\pi t} }}{e^{ - \frac{{{x^2}}}{{2t}}}}dx = 1} $. Let $f \in {L^\infty }\left( \Bbb{R} \right)$ and $u\left( {t,x} \right) = f \star {p_t}\left( x \right)$ where "$\star$" denotes convolution, i.e. $u\left( {t,x} \right) = f \star {p_t}\left( x \right) = \mathop \smallint \limits_R f\left( y \right){p_t}\left( {x - y} \right)dy$. Show that
a) $\frac{{\partial u\left( {t,x} \right)}}{{\partial t}} = \int_R {f\left( y \right)\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} dy,t > 0,x \in \Bbb{R}$
b) $\frac{{\partial u\left( {t,x} \right)}}{{\partial x}} = \int_R {f\left( y \right)\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial x}}} dy,t > 0,x \in \Bbb{R}$

I am able to show a) by the following proof. It is a bit long and follows my professor's lecture notes. The problem is the method seems to fail for b). I have two questions.

*

*Is it possible to make the method work for b)?


*The method looks super lengthy and complicated in calculation. Is there any easier way solve the problem?
PART 1. $\frac{{\partial u\left( {t,x} \right)}}{{\partial t}} = \mathop {\lim }\limits_{h \to 0} \int_R {f\left( y \right)\frac{{{p_{t + h}}\left( {x - y} \right) - {p_t}\left( {x - y} \right)}}{h}} dy$. By mean value theorem $\exists {\xi _{t,h}} \in \left( {t,t + h} \right)$ st.$\frac{{{p_{t + h}}\left( {x - y} \right) - {p_t}\left( {x - y} \right)}}{h} = \left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}$ and $\frac{{\partial u\left( {t,x} \right)}}{{\partial t}} = \mathop {\lim }\limits_{h \to 0} \mathop \smallint \limits_R f\left( y \right)\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}dy$. The subscripts of ${\xi _{t,h}}$ means it is dependent on $t$ and $h$. Further, $\mathop {\lim }\limits_{h \to 0} {\xi _{t,h}} = t$.
Now if $f\left( y \right)\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}$ meets the DCT condition, i.e. $\forall x\forall t > 0\forall h \to 0$ $\exists \phi \left( y \right) \in {L^1}$ st. $\left| {f\left( y \right)\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}} \right| \le \phi \left( y \right)$ , it follows that $\mathop {\lim }\limits_{h \to 0} \mathop \smallint \limits_R f\left( y \right)\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}dy = \mathop \smallint \limits_R \mathop {{\rm{lim}}}\limits_{h \to 0} f\left( y \right)\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}dy = \mathop \smallint \limits_R \mathop {{\rm{lim}}}\limits_{h \to 0} f\left( y \right)\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}dy$ and we are done.
PART2. Since $|f(y)|≤C$ for some constant $C$ a.e., then we only need to show $\exists \phi \left( y \right) \in {L^1}$ st. $\left| {\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}} \right| \le \phi \left( y \right)$. Expanding $\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}$ and we have $\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}} = \frac{1}{{2\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - y} \right)}^2}}}{{2{\xi _{t,h}}}}}}\left[ {{{\left( {x - y} \right)}^2}{\xi _{t,h}}^{ - \frac{5}{2}} - {\xi _{t,h}}^{ - \frac{3}{2}}} \right]$.
Since ${\xi _{t,h}} \in \left( {t,t + h} \right)$ and $h→0$, we can choose a fixed value $0<δ<1$ st. $t - \delta>0$ and ${\xi _{t,h}} \in \left( {t - \delta ,t + \delta } \right)$ holds when $|h|$ is sufficiently small. This step is used in the next part to make $\phi(y)$ independent from $h$.
PART3. Now $∀x$ find a positive N that is large enough st. ${\left( {x - y} \right)^2} \le {e^{\frac{{{{\left( {x - y} \right)}^2}}}{2}}}$ when $\left| y \right| > N$. Then we have $\left| {\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}} \right| \le \frac{1}{{2\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - y} \right)}^2}}}{{2\left( {t + \delta } \right)}}}}\left( {\left| {{{\left( {x - y} \right)}^2}{{\left( {t - \delta } \right)}^{ - \frac{5}{2}}}} \right| + \left| {{{\left( {t - \delta } \right)}^{ - \frac{3}{2}}}} \right|} \right) \le \frac{1}{{2\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - y} \right)}^2}}}{{2\left( {t + \delta } \right)}}}}\left( {\left| {{e^{\frac{{{{\left( {x - y} \right)}^2}}}{2}}}{{\left( {t - \delta } \right)}^{ - \frac{5}{2}}}} \right| + \left| {{{\left( {t - \delta } \right)}^{ - \frac{3}{2}}}} \right|} \right) = \frac{1}{{2\sqrt {2\pi } }}{e^{\frac{{ - {{\left( {x - y} \right)}^2} + t + \delta }}{{2\left( {t + \delta } \right)}}}}\left| {{{\left( {t - \delta } \right)}^{ - \frac{5}{2}}}} \right| + \frac{1}{{2\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - y} \right)}^2}}}{{2\left( {t + \delta } \right)}}}}\left| {{{\left( {t - \delta } \right)}^{ - \frac{3}{2}}}} \right|$ which is obviously Lebesgue integrable $\forall x\forall t > 0\forall 0 < \delta  < 1$ with respect to $y$ on $|y|>N$.
For $|y|≤N$, $\left| {\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}} \right| \le \frac{1}{{2\sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - y} \right)}^2}}}{{2\left( {t + \delta } \right)}}}}\left( {\left| {{{\left( {x - y} \right)}^2}{{\left( {t - \delta } \right)}^{ - \frac{5}{2}}}} \right| + \left| {{{\left( {t - \delta } \right)}^{ - \frac{3}{2}}}} \right|} \right)$. The right-hand side of the inequality is a continuous function with respect to $y$ on a bounded interval, so it is bounded. Thus $∀|y|≤N$, $\left| {\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}} \right| \le M$ for some finite positive constant $M$. A finite constant on an bounded interval is Lebesgue integrable.
Finally, we have proved $f\left( y \right)\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}$ meets the DCT condition and hence $\frac{{\partial u\left( {t,x} \right)}}{{\partial t}} = \mathop \smallint \limits_R f\left( y \right)\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}dy$.

I have expected the same thing can work smoothly for b) but in fact I got stuck.
PART1.  $\frac{{\partial u\left( {t,x} \right)}}{{\partial x}} = \mathop {\lim }\limits_{h \to 0} \mathop \smallint \limits_R f\left( y \right)\frac{{{p_t}\left( {x + h - y} \right) - {p_t}\left( {x - y} \right)}}{h}dy$. By mean value theorem $\exists {\xi _{x,h}} \in \left( {x,x + h} \right)$ st. $\frac{{{p_t}\left( {x + h - y} \right) - {p_t}\left( {x - y} \right)}}{h} = \left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial x}}} \right){|_{x = {\xi _{x,h}}}}$ and $\frac{{\partial u\left( {t,x} \right)}}{{\partial x}} = \mathop {\lim }\limits_{h \to 0} \mathop \smallint \limits_R f\left( y \right)\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial x}}} \right){|_{x = {\xi _{x,h}}}}dy$.
PART2.  Expanding $\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial X}}} \right){|_{x = {\xi _{x,h}}}}$ and we have $\left( {\frac{{\partial {p_x}\left( {x - y} \right)}}{{\partial x}}} \right){|_{x = {\xi _{x,h}}}} =  - \frac{{{\xi _{x,h}} - y}}{{\sqrt {2\pi } {t^{\frac{3}{2}}}}}{e^{ - \frac{{\xi _{x,h}^2}}{{2t}}}}$.
Now the same technique for a) fails for b). We can still choose $\delta$ st. ${\xi _{x,h}} \in \left( {x - \delta ,x + \delta } \right)$ holds when $|h|$ is sufficiently small. It is either $|x-\delta|<|x+\delta|$ or $|x-\delta|>|x+\delta|$. WLOG suppose $|x-\delta|<|x+\delta|$, and we have $\left| {\left( {\frac{{\partial {p_t}\left( {x - y} \right)}}{{\partial t}}} \right){|_{t = {\xi _{t,h}}}}} \right| \le \left| {\frac{{x + \delta  - y}}{{\sqrt {2\pi } {t^{\frac{3}{2}}}}}{e^{ - \frac{{{{\left( {x - \delta } \right)}^2}}}{{2t}}}}} \right|$. However, this time the right-hand side looks not integrable with respect to $y$.
Thank you so much for having patience reading this long post!
 A: I'll be honest, I read the problem statement but did not attempt to read what followed in every detail because it looked like you were trying to reinvent the wheel from first principles. It seems that you are having trouble justifying differentiating under the integral sign. Here is reasonably general theorem you can apply that will help you to not have to reinvent the wheel every time (I have it stored in my notes so that I can refer to it whenever I need it):$\newcommand{\R}{\mathbb{R}}\renewcommand{\d}{\partial}$
Theorem. Let $\phi:[a,b]\times\R^d\to\R$ be continuous. Suppose that for each $y\in\R^d$, $\phi(x,y)$ is differentiable on $[a,b]$. Suppose that for each $x\in[a,b]$, $\phi(x,y)$ is integrable over $\R^d$. Suppose that $\frac{\d\phi}{\d x}(x,y)$ is continuous on $[a,b]\times\R^d$ and is dominated by a non-negative integrable function $f:\R^d\to\R$ uniformly in $x$. Then
$$\frac{d}{dx}\int_{\R^d} \phi(x,y)\,dy = \int_{\R^d} \frac{\d\phi}{\d x}(x,y)\,dy$$
for each $x\in[a,b]$.
Proof. Fix $x\in[a,b]$. For sufficiently small $h$, the difference quotient
$$\frac{\phi(x+h,y)-\phi(x,y)}{h}$$
is defined and, by the mean value theorem, equals $\frac{\d\phi}{\d x}(\xi(x,y,h),y)$ for some $\xi(x,y,h)$ with $|x-\xi(x,y,h)|<|h|$.
Fix $\epsilon>0$ and choose a closed cube $C\subset\R^d$ centered at the origin large enough such that $\int_{\R^d\setminus C} f(y)\,dy<\epsilon/3$. Then choose $\delta>0$ such that
$$\left|\frac{\d\phi}{\d x}(x,y)-\frac{\d\phi}{\d x}(\xi,y)\right|<\frac{\epsilon}{3\operatorname{vol}(C)}$$
whenever $|x-\xi|<\delta$ and $y\in C$. This is possible because $\frac{\d\phi}{\d x}$ is uniformly continuous on $[a,b]\times C$. Then when $|h|<\delta$ and $y\in C$,
$$\left|\frac{\phi(x+h,y)-\phi(x,y)}{h}-\frac{\d\phi}{\d x}(x,y)\right| < \frac{\epsilon}{3\operatorname{vol}(C)}$$
and thus
\begin{align*}
&\left|\frac{\int_{\R^d} \phi(x+h,y)\,dy - \int_{\R^d} \phi(x,y)\,dy}{h} - \int_{\R^d}\frac{\d\phi}{\d x}(x,y)\,dy\right| \\
&\qquad\leq \int_{y\in\R^d\setminus C} + \int_{y\in C} \left|\frac{\phi(x+h,y)-\phi(x,y)}{h}-\frac{\d\phi}{\d x}(x,y)\right|\,dy \\
&\qquad< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} \\
&\qquad = \epsilon.
\end{align*}
This completes the proof.
Phew! OK, now that we have the wheel ready for us to use, let's use it. In (b), $d=1$ and we want to apply this theorem to $\phi(x,y)=f(y)p_t(x-y)$ with $t>0$ fixed. Let's let $[a,b]=[-N,N]$. All the hypotheses check out fine except we need to make sure that
$$\frac{\d\phi}{\d x}=f(y)\frac{\d p_t}{\d x}(x-y)$$ is dominated by an integrable function uniformly for $x$ in $[-N,N]$. Since $f$ is bounded, it's enough to show that $\frac{\d p_t}{\d x}(x-y)$ is dominated by an integrable function uniformly for $x\in[-N,N]$. We know that $\frac{\d p_t}{\d x}(0-y)$ is integrable; this just says that that function and all its translates to the left or right by a distance less than or equal to $N$ lie beneath one single integrable function. Seems reasonable, and yes, it's true.
To show this, since $t$ is fixed I'm just going to assume $p_t(x)=e^{-x^2}$; adding in the constants just makes what I'm about to write down look messier and obscures the idea. We then have
$$\frac{\d p_t}{\d x}(x-y)=-2(x-y)e^{-(x-y)^2},$$
and again, we want to show that this function is dominated by an integrable function, uniformly for $x\in[-N,N]$. When $|y|\geq 2N$, we have $|x|\leq N\leq |y|/2$ and thus
$$|x-y|\geq |y|-|x|\geq |y|-\frac{|y|}{2} = \frac{|y|}{2}.$$
Therefore when $|y|\geq 2N$,
$$|-2(x-y)e^{-(x-y)^2}| \leq 2(N+|y|)e^{-|y|^2/4}.$$
This function is integrable! When $|y|\leq 2N$, the function we're examining is bounded uniformly in $x$ by $6N$. Thus the function given by $6N$ when $|y|\leq 2N$ and $2(N+|y|)e^{-|y|^2/4}$ when $|y|>2N$ is (1) integrable and (2) dominates $\frac{\d p_t}{\d x}(x-y)$ uniformly for $x\in[-N,N]$. And that means we're done!
We've shown the result holds for all $x\in[-N,N]$. But since $N$ was arbitrary, the result holds for all $x\in\R$.
I'm sorry if my answer was a little roundabout, but when thinking about differentiating under the integral I like to have a big theorem that I can use, and I never want to go all the way back to the mean value theorem.
