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for some positive integer $d \geq 1$ I have a globally Lipschitz continuous function $f \colon \mathbb{R}^d \to \mathbb{R}$ with Lipschitz constant $1$ and would like to approximate it by a sequence $f_\varepsilon$ with the following properties:

  1. For all $\varepsilon > 0$, the function $f_\varepsilon$ is $k$ times partially differentiable and all its partial derivatives up to order $k$ are bounded. Here, $k$ is some positive integer ($k=3$ suffices for my purposes but arbitrary $k$ would be nicer);
  2. For all $\varepsilon > 0$, it holds that $\| f_\varepsilon \|_{\text{Lip}} \leq \| f \|_{\text{Lip}}$, where $\|\cdot\|_\text{Lip}$ denotes the usual Lipschitz norm;
  3. For $\varepsilon \to 0$, it holds that $\| f_\varepsilon - f \|_{\infty} \to 0$.

For $d=1$ and $k=2$ I have found the following example in the literature which is stated without proof (it should continue to work for arbitrary $d$ if one replaces the one dimensional Gaussian distribution by a $d$-dimensional one) but don't understand why the second derivative exists (in the classical sense) and is bounded: $$f_\varepsilon(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} g(x + \varepsilon y) e^{- y^2/2} \, \mathrm{d}y.$$ Can anybody give me a hint or provide an example of such a sequence?

Thanks in advance!

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  • $\begingroup$ Do you want the (2nd and higher) derivatives to be bounded for each $\epsilon$ by a constant that may depend on $\epsilon$ or uniformly in $\epsilon$? If uniformly, this is not possible in general. $\endgroup$ – PhoemueX Sep 19 '14 at 21:09
  • $\begingroup$ @PhoemueX In my case, the derivatives can be bounded by a constant depending on $\varepsilon$ but that's a very interesting point. Is it already impossible in one dimension? What about the norms $\| \partial (f_{\varepsilon} - f)\|_{\infty}$, where $\partial$ stands for some partial derivatives, can these be bounded uniformly in $\varepsilon$? $\endgroup$ – herrsimon Sep 20 '14 at 7:36
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Let $$ h(x)=\left\{\begin{array}{lll}\mathrm{e}^{-1/x^2} & \text{if} & x>0,\\ 0 & \text{if} & x\le 0.\end{array}\right. $$ Then $h\in C^\infty(\mathbb R)$. Then set $$ j(\boldsymbol{x})=c\,h\big(1-\|\boldsymbol{x}\|^2\big), $$ where $\boldsymbol{x}\in\mathbb R^n$, and $c>0$, so that $\int_{\mathbb R^n}j(\boldsymbol{x})\,d\boldsymbol{x}=1$. Clearly, $j\ge 0$, $j\in C^\infty(\mathbb R^n)$ and $\,\mathrm{supp}\,j\subset B(0,1)$ - the unit ball.

Next define $j_e(\boldsymbol{x})=\varepsilon^{-n}j(\varepsilon^{-1}\boldsymbol{x})$. Then $\int_{\mathbb R^n}j_\varepsilon(\boldsymbol{x})\,d\boldsymbol{x}=1$ and let the function $$ f_\varepsilon=f*j_\varepsilon, $$ i.e., $$ f_\varepsilon(\boldsymbol{x})=\int_{\mathbb R^n} f(\boldsymbol{y})\,j_\varepsilon(\boldsymbol{x}-\boldsymbol{y})\,d\boldsymbol{y}=\int_{\mathbb R^n} f(\boldsymbol{x}-\boldsymbol{y})\,j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y}= \frac{1}{\varepsilon^n}\int_{B(0,\varepsilon)} f(\boldsymbol{x}-\boldsymbol{y})\,j(\boldsymbol{y}/\varepsilon)\,d\boldsymbol{y}=\frac{1}{\varepsilon^n}\\=\int_{B(0,1)} f(\boldsymbol{x}-\varepsilon\boldsymbol{y})\,j(\boldsymbol{y})\,d\boldsymbol{y}. $$ Clearly $f_\varepsilon\in C^\infty(\mathbb R^n)$. Next $$ f_\varepsilon(\boldsymbol{x}_1)-f_\varepsilon(\boldsymbol{x}_2)= \int_{\mathbb R^n} \big(f(\boldsymbol{x}_1-\boldsymbol{y})-f(\boldsymbol{x}_2-\boldsymbol{y})\big)\,j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y} $$ and hence $$ \lvert\,f_\varepsilon(\boldsymbol{x}_1)-f_\varepsilon(\boldsymbol{x}_2)\rvert\le \int_{\mathbb R^n} \lvert\, f(\boldsymbol{x}_1-\boldsymbol{y})-f(\boldsymbol{x}_2-\boldsymbol{y})\rvert\,j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y}\le\|\boldsymbol{x}_1-\boldsymbol{x}_2\|\int_{\mathbb R^n}j_\varepsilon(\boldsymbol{y})\,d\boldsymbol{y}=\|\boldsymbol{x}_1-\boldsymbol{x}_2\|. $$ Finally $$ \lvert\,f_\varepsilon(\boldsymbol{x})-f(\boldsymbol{x})\rvert\le \left|\int_{B(0,1)} \big(f(\boldsymbol{x}-\varepsilon\boldsymbol{y})-f(\boldsymbol{x})\big)\,j(\boldsymbol{y})\,d\boldsymbol{y}.\,\right|\le \cdots\le \varepsilon. $$

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  • $\begingroup$ Thank you, now I've understood the main arguments. Actually, you still have to show that all (partial) derivatives are bounded, right? Although this is heuristically clear, I think that taking $\exp{x^2}$ instead of $\exp{1/x^2}$ from the start as in my one-dimensional example would have simplified things (unless I'm missing something). $\endgroup$ – herrsimon Sep 20 '14 at 7:26
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Rewrite $f_{\epsilon}(x)$ by substituting $y=y'-\frac{1}{\epsilon}x$: $$ \begin{align} f_{\epsilon}(x) & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\epsilon y') e^{-(y'-\frac{1}{\epsilon}x)^{2}/2}\,dy' \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\epsilon y')e^{-(\epsilon y'-x)^{2}/2\epsilon^{2}}\,dy' \end{align} $$ Now you can see that the derivatives respect to $x$ fall onto the Gaussian, instead onto the function $g$. You can further substitute $y''=\epsilon y$ to make the integral a little more transparent.

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  • $\begingroup$ And again it's a shame that one can only accept a single answer... thanks for the clarification, so $f_{\varepsilon}$ is indeed $\mathcal{C}^{\infty}$. $\endgroup$ – herrsimon Sep 20 '14 at 7:05

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