$L^1$ norm of convolution Let $f_{\lambda} = \frac{\lambda}{2}e^{-\lambda |x|}$. Prove that $||f_{\lambda} \ast g - g || \to 0$, when $\lambda \to \infty$, where $g \in L^1$
 A: Note that for the characteristic function of an interval, it is simple to prove that
$$
\lim_{h\to0}\int_{\mathbb{R}}|\chi_I(x+h)-\chi_I(x)|\,\mathrm{d}x=0\tag{1}
$$
Approximating simple functions by a finite linear combination of such characteristic functions, then Lebesgue integrable functions by simple functions, we can show that
$$
\lim_{h\to0}\int_{\mathbb{R}}|g(x+h)-g(x)|\,\mathrm{d}x=0\tag{2}
$$
for all Lebesgue integrable $g$.

For all $\lambda\gt0$,
$$
\int_{\mathbb{R}}f_\lambda(x)\,\mathrm{d}x=1\tag{3}
$$
Furthermore, for any $h\gt0$,
$$
\int_{|x|\gt h}|f_{\lambda}(x)|\,\mathrm{d}x=e^{-\lambda h}\tag{4}
$$
Next is simply use of $(2)$, $(3)$, and $(4)$:
$$
\begin{align}
\int_{\mathbb{R}}|f_\lambda\ast g(x)-g(x)|\,\mathrm{d}x
&=\int_{\mathbb{R}}\int_{\mathbb{R}}|f_\lambda(y)(g(x-y)-g(x))|\,\mathrm{d}y\,\mathrm{d}x\\
&\le\int_{\mathbb{R}}\int_{|y|\le h}|f_\lambda(y)(g(x-y)-g(x))|\,\mathrm{d}y\,\mathrm{d}x\\
&+\int_{\mathbb{R}}\int_{|y|\le h}|f_\lambda(y)(g(x-y)-g(x))|\,\mathrm{d}y\,\mathrm{d}x\\
&\le\|f_\lambda\|_{L^1}\sup_{|y|\le h}\|g(x-y)-g(x)\|_{L^1(x)}\\
&+\|f_\lambda(x)\|_{|x|\gt h}2\|g\|_{L^1}\\
&\le\sup_{|y|\le h}\|g(x-y)-g(x)\|_{L^1(x)}+2e^{-\lambda h}\|g\|_{L^1}\tag{5}
\end{align}
$$
By choosing $h$ small enough so that $\sup_{|y|\le h}\|g(x-y)-g(x)\|_{L^1(x)}\le\frac\epsilon2$, then $\lambda$ big enough so that $2e^{-\lambda h}\|g\|_{L^1}\le\frac\epsilon2$, we get
$$
\int_{\mathbb{R}}|f_\lambda\ast g(x)-g(x)|\,\mathrm{d}x\le\epsilon\tag{6}
$$
