Problem: Let $f \in L ^ { 1 } , | \widehat { f } | \in L ^ { 1 }$,$$
u ( x , t ) = \int _ { – \infty } ^ { + \infty } \widehat { f } ( \xi ) e ^ { 2 \pi i \xi x – 4 \pi ^ { 2 } a ^ { 2 } \xi ^ { 2 } t } d \xi
$$Show that $$
\lim _ { t \rightarrow 0 } u ( x , t ) = f ( x ) \text { uniformly in } x
$$
In the previous question, we showed that $$
\left\{ \begin{array} { c c } { \frac { \partial u } { \partial t } = a ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } } & { : x \in \mathbb { R } , t > 0 } \\ { u ( x , 0 ) = f ( x ) } & { : x \in \mathbb { R } } \end{array} \right.
$$
Solution: We have that $$
\begin{aligned} | u ( x , t ) – f ( x ) | & = \left| \int _ { \mathbb { R } } \widehat { f } ( \xi ) e ^ { 2 \pi i \xi x } \left( e ^ { – 4 \pi ^ { 2 } a ^ { 2 } \xi ^ { 2 } t } – 1 \right) d \xi \right| \\ & \leq \int _ { \mathbb { R } } | \widehat { f } ( \xi ) | \left| e ^ { – 4 \pi ^ { 2 } a ^ { 2 } \xi ^ { 2 } t } – 1 \right| \end{aligned}
$$
We can see that $$
| \widehat { f } ( \xi ) | \left| e ^ { – 4 \pi ^ { 2 } a ^ { 2 } \xi ^ { 2 } t } – 1 \right| \leq | \widehat { f } ( \xi ) | \in L ^ { 1 } ( \mathbb { R } )
$$
and for any fixed $\xi$ $$
\lim _ { t \rightarrow 0 } | \widehat { f } ( \xi ) | \left| e ^ { – 4 \pi ^ { 2 } a ^ { 2 } \xi ^ { 2 } t } – 1 \right| = 0
$$
and we get uniform convergence with the dominated convergence theorem (DCT).
As I understood, the DCT is a great tool to change limit with integral. However I don't really see what is the relation with uniform convergence. I know that if a sequence $f_n$ converges uniformly to $f$ you can exchange limit and integration, which is also the result of the DCT.
Best Answer
The point is that the inequality $$ | u ( x , t ) - f ( x ) | \leq \int _ { \mathbb { R } } | \widehat { f } ( \xi ) | \left| e ^ { - 4 \pi ^ { 2 } a ^ { 2 } \xi ^ { 2 } t } - 1 \right| $$ holds independently of $x$ and that the term on the RHS converges to $0$ as $t\to 0$. This means that the convergence is uniform in $x$. The Dominated Convergence Theorem is used to prove that RHS converges.
Your remark that uniform convergence of $f_n\to f$ implies the validity of exchanging the limit and integral is not relevant to this problem. In fact, it only holds on a domain with finite measure.