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\vspace*{.3in}
\noindent{\bf Beals, page 34 \#3:}
Let $\epsilon > 0$ be given.  By Theorem 2 on page 33, there is a continuous, 
compactly supported function $g$ such that $\norm[1]{g-f} < \epsilon/4$.  
Note that in fact $\norm[1]{g_a-f_a} = \norm[1]{g-f} < \epsilon/4$ for
all $a\in\R$.

Let $M>0$ be chosen so that $g\equiv 0$ outside $[-M,M]$.  Because
$g$ has compact support, it is uniformly continuous.  So I can choose 
$\delta > 0$ so that
$|x-y|<\delta$ implies that $|g(x) - g(y)| < \epsilon/8M$.  I can, of
course, assume that $\delta<M$, too.  Therefore, if $|a| < \delta$, it
follows that $g_a\equiv 0$ outside $[-2M,2M]$ and 
\begin{eqnarray*}
\norm[1]{f - f_a} 
& \leq & 
\norm[1]{f - g} + \norm[1]{g - g_a} + \norm[1]{f_a - g_a} \\
& < & \frac{\epsilon}{4} + \int_{-2M}^{2M} (g(x) - g(x-a)) 
+ \frac{\epsilon}{4} \\
& < & \frac{\epsilon}{2} + \int_{-2M}^{2M} \frac{\epsilon}{8M} = \epsilon.
\end{eqnarray*}
This proves that $\lim_{a\to 0} \norm[1]{f-f_a} = 0$.
\qed
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