# pointwise approximation to the identity

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!

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.
• But this is not what I have$|\int_{\{y:|y|\leq \Delta\}} f(x-\epsilon y)-f(y)]d\mu (y)|$, I have $$|\int_{\{y:|y|\leq \Delta\}} f(x-\epsilon y)-f(x)]d\mu (y)|$$ Note the $x$ in the second $f$. Then I do not see why I can interchange limits as $\Delta\rightarrow\infty$ and $\varepsilon\rightarrow 0$, because then I have $$\left|\int_{\{y:|y|\leq \Delta\}} f(x-\epsilon y)-f(x)]d\mu (y)\right|\leq \varepsilon L\int_{\{y:|y|\leq \Delta\}}|y|d\mu (y)$$ – Pink Panther Apr 7 at 13:57
Instead of splitting the integral at $$|y|=\frac{1}{\varepsilon}$$, we can do it at $$|y|=\frac{1}{\sqrt\varepsilon}$$. Then we get $$\int_{|y|<\frac{1}{\sqrt\varepsilon}}L|\varepsilon y|\mathrm{d}\mu(y) =\varepsilon L\int_{|y|<\frac{1}{\sqrt\varepsilon}}|y|\mathrm{d}\mu(y) \leq\sqrt\varepsilon L\int_{|y|<\frac{1}{\sqrt\varepsilon}}\mathrm{d}\mu(y) \xrightarrow{\varepsilon\rightarrow 0}0$$