Using approximation of a function $f \in L^1 \cap L^\infty$. Let $f \in L^1(m) \cap L^\infty(m)$.  Define a functions $\varphi$ on $\mathcal{R}$ by letting $\varphi(x) := \int f(x-t)f(t)dm(t)$.
(a)  Prove that the integral defining $\varphi$ is well defined.  That is, for each $x \in \mathcal{R}$, the function $t \to f(x-t)f(t)$, is in $L^1(m)$.
(b) Prove that $\varphi$ is continuous.
(c) Prove that $\lim_{\vert x\vert \to \infty} \varphi(x)$ exists.  Find it.
For part (a), I know that $\int \vert f(x-t)f(t) \vert dm(t) \leq \int \vert\vert f\vert\vert_\infty dm(t)$, but I cannot see why that is finite.
For part (b), I know that for any $\epsilon > 0$, there exists a continuous function $g$ with compact support such that $\int \vert f-g\vert < \epsilon$ and I have $\vert \varphi(x) - \varphi(y) \vert \leq \int \vert f(x-t)f(t) - f(y-t)f(t)\vert dm(t)$.  I want to replace $f$ in this equation with $g$, but I am not sure how to go about that.
For part (c), again I want to use the the approximation $g$, as I know $\lim_{\vert x\vert \to \infty} g(x) = 0.$
Any suggestions are welcome. 
 A: For (a), you missed an $f$. We have
$$ \def\abs#1{\left|#1\right|}\def\R{\mathbb R}\int_\R \abs{f(t)f(x-t)}\, dm(t) \le \def\norm#1{\left\|#1\right\|}\norm f_\infty \int_\R \abs{f(t)}\, dt = \norm f_\infty\norm f_1 < \infty $$
For (b) use the dominated convergence theorem. As $t \mapsto f(t)f(x-t)$ is dominated by $f \cdot \norm f_\infty$independent of $x$, we have for $x_n \to x$ that 
$$ \phi(x_n) = \int_\R f(x_n- t)f(t)\, dt \to \int_\R f(x-t)f(t)\, dt = \phi(x) $$
For (c), given $\epsilon > 0$, choose an $M$ such that $\int_{\abs t \ge M} \abs{f(t)}\, dt \le \epsilon/2\norm f_\infty$. For $\abs x \ge 2M$ we have 
\begin{align*}
  \abs{\phi(x)} &\le \int_\R \abs{f(t)}\abs{f(x-t)}\, dt\\
                &\le \norm f_\infty \int_{\abs t \ge M} \abs{f(t)}\, dt
               + \norm f_\infty \int_{\abs t \le M} \abs{f(x-t)}\, dt\\
                &\le 2\norm f_\infty \int_{\abs t\ge M} \abs{f(t)}\, dt \quad \text{as $\abs x \ge 2M$ and $\abs t \le M$ imply $\abs{x-t}\ge M$} \\
                &\le \epsilon
\end{align*}
