pointwise approximation to the identity

Question

Let $\mu$ be a positive Borel measure on $\mathbb{R}$ with $\int_{\mathbb{R}}\mathrm{d}\mu=1$. For $\varepsilon>0$ define the measures $\mu_\varepsilon$ by $$\int f\mathrm{d}\mu_\varepsilon=\int f(\varepsilon y)\mathrm{d}\mu(y)$$

I have to show the following things

a) $\mu_x*f(x)\rightarrow f$ pointwise as $\varepsilon\rightarrow 0$ for every bounded Lipschitz function.
b) Assume that in addition $\int_{\mathbb{R}}|x|\mathrm{d}\mu(x)<\infty$. Show that $\mu_\varepsilon * f\rightarrow f$ pointwise as $\varepsilon\rightarrow 0$ for every Lipschitz function.

Attempt to prove a) Let $f$ be a bounded Lipschitz function such that $\|f\|_\infty=M<\infty$ and let $L\in(0,\infty)$ such that $|f(x)-f(y)|\leq L|x-y|$ for all $x,y\in\mathbb{R}$. Let $\varepsilon>0,\, x\in\mathbb{R}$ and let $f_\varepsilon(x)=\int_{\mathbb{R}}f(x-y)\mathrm{d}\mu_\varepsilon(y)$. Then \begin{align*} |f_\varepsilon(x)-f(x)| &=\left|\int_{\mathbb{R}}f(x-\varepsilon y)\mathrm{d}\mu(y) -\int_{\mathbb{R}}f(x)\mathrm{d}\mu\right| \\ &\leq\left|\int_{|y|<\frac{1}{\varepsilon}}f(x-\varepsilon y)-f(x)\mathrm{d}\mu(y)\right| +\left|\int_{|y|\geq\frac{1}{\varepsilon}}f(x-\varepsilon y)-f(x)\mathrm{d}\mu(y)\right| \\ &\leq\int_{|y|<\frac{1}{\varepsilon}}|f(x-\varepsilon y)-f(x)|\mathrm{d}\mu(y) +\int_{|y|\geq\frac{1}{\varepsilon}}|f(x-\varepsilon y)-f(x)|\mathrm{d}\mu(y) \\ &\leq\int_{|y|<\frac{1}{\varepsilon}}L|x-\varepsilon y-x|\mathrm{d}\mu(y) +2M\int_{|y|\geq\frac{1}{\varepsilon}}\mathrm{d}\mu(y) \end{align*} We have $$\int_{|y|<\frac{1}{\varepsilon}}L|\varepsilon y|\mathrm{d}\mu(y) =\varepsilon L\int_0^1\int_{|y|<\frac{1}{\varepsilon}}t|y|\mathrm{d}\mu(y)\mathrm{d}t\overset ?\rightarrow 0 $$ Moreover $$\int_{|y|\geq\frac{1}{\varepsilon}}\mathrm{d}\mu(y) =\int_{\mathbb{R}}\mathrm{d}\mu-\int_{|y|<\frac{1}{\varepsilon}}\mathrm{d}\mu \xrightarrow{\varepsilon\rightarrow 0}0$$ Therefore $$|f_\varepsilon(x)-f(x)| \leq\int_{|y|<\frac{1}{\varepsilon}}L|x-\varepsilon y-x|\mathrm{d}\mu(y) +2M\int_{|y|\geq\frac{1}{\varepsilon}}\mathrm{d}\mu(y) \xrightarrow{\varepsilon\rightarrow 0}0$$

As you can see, I am almost done, but I am not sure about how to show that the part with $?$ above the arrow.
I thought about substituting somehow, but I am kinda clueless here...

For part b) I had the following idea, but it somehow seems too straightforward to me.

Proof of b): Let $f$ be Lipschitz and $L>0$ such that $|f(x)-f(y)|\leq L|x-y|$.
Let $\varepsilon>0,\, x\in\mathbb{R}$ and let $f_\varepsilon(x)=\int_{\mathbb{R}}f(x-y)\mathrm{d}\mu_\varepsilon(y)$. As above, we have \begin{align*} |f_\varepsilon(x)-f(x)| &\leq\int_{\mathbb{R}}|f(x-\varepsilon y)-f(x)|\mathrm{d}\mu(y) \\ &\leq L\int_{\mathbb{R}}|\varepsilon y|\mathrm{d}\mu(y) \\ &=\varepsilon L\int_{\mathbb{R}}|y|\mathrm{d}\mu(y) \end{align*} Hence $|f_\varepsilon(x)-f(x)|\xrightarrow{\varepsilon\rightarrow 0}0$, because $\int_{\mathbb{R}}|x|\mathrm{d}\mu(x)<\infty$.

To me this seems too simple, but I'm pretty sure this works.

Any help would be very much appreciated!

Answer 1

I found a solution.
Instead of splitting the integral at $|y|=\frac{1}{\varepsilon}$, we can do it at $|y|=\frac{1}{\sqrt\varepsilon}$. Then we get $$\int_{|y|<\frac{1}{\sqrt\varepsilon}}L|\varepsilon y|\mathrm{d}\mu(y) =\varepsilon L\int_{|y|<\frac{1}{\sqrt\varepsilon}}|y|\mathrm{d}\mu(y) \leq\sqrt\varepsilon L\int_{|y|<\frac{1}{\sqrt\varepsilon}}\mathrm{d}\mu(y) \xrightarrow{\varepsilon\rightarrow 0}0 $$

Answer 2

Your argument requires some modifications. Given $\eta >0$ choose $\Delta$ such that $\mu \{y:|y| >\Delta\} <\eta$. Note that $|\int_{\{y:|y|> \Delta\}} f(x-\epsilon y)-f(y)]d\mu (y)|\leq 2M \mu \{y:|y| >\Delta\} <2M\eta$. Now use Lipschitz condition for $|\int_{\{y:|y|\leq \Delta\}} f(x-\epsilon y)-f(y)]d\mu (y)|$. Can you complete the argument now? The second part is easier.