Assume that $f:[a,b]\rightarrow \mathbb R$ is continuous and $g:\mathbb R\rightarrow \mathbb R$ is Lipschitz continuous. Now define $$ h(x):=\int_a^b g(t-x)f(t)dt. $$ I want to show that $h$ is Lipschitz continuous.
My attempt of proof is:
Since $g$ is Lipschitz continuous, then there exists $K>0$ such that $$|g(x)-g(y)|\leq K |x-y|.$$ In particular we have \begin{eqnarray} |g(t-x)-g(t-y)|\leq K |t-x-(t-y)|=K|x-y| \qquad \qquad (1) \end{eqnarray} and multiplying by $|f(t)|$ we obtain $$\big|g(t-x)-g(t-y)\big|\big|f(t)\big| = \big|\left[g(t-x)-g(t-y)\right] f(t)\big|\leq K |f(t)||x-y|$$ Since $g$ is also continuous then $g$ is Riemann integrable on $[a,b]$ and thus \begin{eqnarray*} \left|h(x)-h(y)\right| &=& \left|\int_a^b g(t-x)f(t)dt-\int_a^b g(t-y)f(t)dt\right| \\ &=& \bigg|\int_a^b\left[g(t-x)-g(t-y)\right] f(t)dt\bigg| \\ &\leq & \int_a^b \big|\left[g(t-x)-g(t-y)\right] f(t)\big|dt\\ &\leq & \int_a^b K |f(t)||x-y|dt\\ &=& \mathcal K|x-y| \end{eqnarray*} where $\mathcal K= K\int_a^bf(t)dt$ is a real constant. Hence $h$ is Lipschitz continuous. $\blacksquare$
My concern is about the use of inequality (1). I am not sure if I can do that. Do I need to specify the value of $t$? For example, mentioning that $t\in[a,b]$.
If you have some advice and feedback about this proof (maybe it is incorrect), I would appreciate it. Thanks.
Edit: Changed "$K\in \mathbb R$" for $K>0$ in the first line.
