$L^1$ norm of convolution

Question

Let $f_{\lambda} = \frac{\lambda}{2}e^{-\lambda |x|}$. Prove that $||f_{\lambda} \ast g - g || \to 0$, when $\lambda \to \infty$, where $g \in L^1$

Answer 1

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 $g$.

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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 $$ \int_{\mathbb{R}}|f_\lambda\ast g(x)-g(x)|\,\mathrm{d}x\le\epsilon\tag{6} $$