Exercise 8.4.19

Let’s first show that $e^{-\mathit{\alpha t}}$ is continuous. For any fixed ${\alpha }_0$ and $t_0$, we want $\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0$, that whenever $\parallel (\alpha ,t)-({\alpha }_0,t_0)\parallel <\delta >0$, that

Since $e^{-x}$ is continuous, we can find ${\delta }_1$ so that

Now if $\left|t_0-t\right|<\frac{{\delta }_1}{2\left|{\alpha }_0\right|}={\delta }_t$, $\left|t_0-t\right|<1$, and $\left|{\alpha }_0-\alpha \right|<\frac{{\delta }_1}{2(\left|t_0\right|+1)}={\delta }_{\alpha }$, then

We can ensure both of these conditions by setting $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->\{{\delta }_t,{\delta }_{\alpha },1\}$, thus showing $f(\alpha ,t)$ is continuous.

It’s fairly easy to show that if $g(x)$ is continuous, so is $g(x,y)=g(x)$. Additionally, a careful re-reading of the proofs for Theorem 4.2.3 (Sequential Criterion for Functional Limits) and Exercise 4.2.1 (Algebraic Limit Theorem for Functional Limits) show that they can be reused for complete metric spaces such as $R^2$ with the usual metric. Thus Theorem 4.3.4 (Algebraic Continuity Theorem) can also be generalized to $R^2$, and we can conclude for any $x\in R$, the function $g(\alpha ,t)=t^x{e}^{-\mathit{\alpha t}}$ is continuous. Thus, both $f$ and $f_{\alpha }$ are continuous.

Now we need to show ${\int }_0^{\infty }{f}_{\alpha }(\alpha ,t)\mathit{dt}$ converges uniformly when $\alpha \ge p>0$. To make part (b) easier we’ll prove the more general statement that ${\int }_0^{\infty }{t}^x{e}^{-\mathit{\alpha t}}\mathit{dt}$ converges uniformly, for any $x\in R$ with $x\ge 0$. As a prerequisite, note that if ${\int }_a^{\infty }g(x,y)\mathit{dy}$ converges uniformly, then so does ${\int }_b^{\infty }g(x,y)\mathit{dy}$, provided ${\int }_a^{b}g(x,y)\mathit{dy}$ is well defined.

We can adapt the solution to Exercise 8.4.5 to show that for any $x\in R$, $\alpha >0$, $\underset{t\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{t}^x{e}^{-\mathit{\alpha t}}=0$. Thus, for any fixed $x\ge 0$, we can find some $M$ so that $g(\alpha )={\int }_M^{\infty }{t}^x{e}^{-\mathit{\alpha t}}\mathit{dt}$ converges uniformly, by Exercise 8.4.16 (variant of the Weierstrauss M-Test) and comparison against $e^{-\mathit{\alpha t}∕2}$. This in turn shows that ${\int }_0^{\infty }{t}^x{e}^{-\mathit{\alpha t}}\mathit{dt}$ converges uniformly.

Finally, before applying Theorem 8.4.9 to get our desired result, we need to define the domain $D$. For a given $\alpha >0$, we want $D$ to contain all points of the form $\alpha ,t$, so define $D=(x,t):\alpha ∕2\le x\le \alpha +1,0\le t$.