Are the partial derivatives of a mollifier bounded?

Question

More precisely, I am wondering whether for any mollifier $ \varphi : \mathbb{R}^n \to \mathbb{R} $ there exists a constant $ M $ such that $ \frac{\partial \varphi}{\partial x^i} (x) \leq M $ for all $ x \in \mathbb{R}^n $. Recall, a mollifier $ \varphi $ is a function which satisfies the following conditions:

1. $ \varphi \in C^{\infty}_{0}(\mathbb{R}^n) $, with $ \mathrm{supp} = \{ x \in \mathbb{R}^n : \lvert x\rvert \leq 1 \} $;
2. $ \varphi > 0 $;
3. $ \int_{\mathbb{R}^n} \varphi = 1 $.

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This question occurred to me whilst trying to prove the following result:

Let $ \varphi : \mathbb{R}^n \to \mathbb{R} $ be a mollifier and define $ \varphi_{\epsilon} := \epsilon^{-n} \varphi(x/\epsilon) $, so that the defining properties of a mollifier listed above automatically imply that $ \varphi_{\epsilon} \in C^{\infty}_{0}(\mathbb{R}^n) $, $ \mathrm{supp} = B(0,\epsilon) $ and $ \int_{\mathbb{R}^n} \varphi_{\epsilon} = 1 $.

Furthermore let $ u: \mathbb{R}^n \to \mathbb{C} $ be a locally integrable function and define $ u_{\epsilon} $ to be the convolution $ u_{\epsilon} := u \ast \varphi_{\epsilon} $.

Then the function $ u_\epsilon : \mathbb{R}^n \to \mathbb{C}$ is $ C^{\infty} $.

My attempt of proving this result begins as follows: Let $ e_i $ be the $i$'th unit vector of the standard basis of $ \mathbb{R}^n $, that is, $ e_i = (e_i^1, ..., e_i^n) $ with $ e_i^j = \delta_{ij} $, where the latter is the Kronecker delta. Then, for a fixed $ x \in \mathbb{R}^n $, we have

\begin{align*} \frac{\partial u_{\epsilon}}{\partial x^i} (x) &= \frac{\partial}{\partial x^i} (u \ast \varphi_{\epsilon}) (x) \\ &= \lim_{h \to 0} \frac{1}{h} \left\{ (u \ast \varphi_{\epsilon})(x + h e_i) - (u \ast \varphi_{\epsilon})(x) \right\} \\ &= \lim_{h \to 0} \int_{\mathbb{R}^n} u(y) \left\{ \frac{\varphi_{\epsilon}(x-y+he_i) - \varphi_{\epsilon}(x-y)}{h} \right\} \mathrm{d}y . \end{align*}

Now the most elegant way to reach the desired conclusion would probably be to use the mean value theorem, by which there must exist a constant $ 0 < c(h) < h $ such that

\begin{align*} \lim_{h \to 0} \int_{\mathbb{R}^n} u(y) \left\{ \frac{\varphi_{\epsilon}(x-y+he_i) - \varphi_{\epsilon}(x-y)}{h} \right\} \mathrm{d}y &= \lim_{h \to 0} \int_{\mathbb{R}^n} u(y) \: \frac{\partial \varphi_{\epsilon}}{\partial x^i} (x-y+c(h)) \: \mathrm{d}y \\ &= \lim_{h \to 0} \left(u \ast \frac{\partial \varphi_{\epsilon}}{\partial x^i}\right) (x+c(h)) . \end{align*}

It is not hard to show that the convolution of a locally integrable function with a continuous function of compact support is continuous and thus we obtain

\begin{align*} \frac{\partial u_{\epsilon}}{\partial x^i} (x) &= \lim_{h \to 0} \left(u \ast \frac{\partial \varphi_{\epsilon}}{\partial x^i}\right) (x+c(h)) \\ &= \left(u \ast \frac{\partial \varphi_{\epsilon}}{\partial x^i}\right) (x) . \end{align*}

I began wondering, however, if there is an alternative to the application of the mean-value theorem. If one could show, for instance, that the integrand $ \left\{ u(y) \left( \frac{\varphi_{\epsilon}(x-y+he_i) - \varphi_{\epsilon}(x-y)}{h} \right) \right\} $ satisfies the conditions of Lebesgue's dominated convergence theorem, then one could interchange the limit and integral in the first equation above to immediately obtain that $ \frac{\partial u_{\epsilon}}{\partial x^i} (x) = \left(u \ast \frac{\partial \varphi_{\epsilon}}{\partial x^i}\right) (x) $. But the difficulty lies in showing that the integrand is dominated by an $ L^1 $ function. However, if we knew that the partial derivatives of $ \varphi $ (and thus of $ \varphi_\epsilon $) were bounded, then the continuity of these partial derivatives would imply (via the mean value theorem I believe...) that for all $ h \leq \delta $, where $ \delta > 0 $ is some constant, there exists a constant $ c > 0 $ such that for all $ y $

\begin{align} \left \lvert \frac{\varphi_{\epsilon}(x-y+he_i) - \varphi_{\epsilon}(x-y)}{h} - \frac{\partial \varphi_{\epsilon}}{\partial x^i}(x-y) \right \rvert \leq c , \end{align}

(remember $ x $ was fixed) and so

\begin{align} \frac{\partial \varphi_{\epsilon}}{\partial x^i}(x-y) - c \leq \frac{\varphi_{\epsilon}(x-y+he_i) - \varphi_{\epsilon}(x-y)}{h} \leq \frac{\partial \varphi_{\epsilon}}{\partial x^i}(x-y) + c . \end{align}

Now just let $ M := \max \left\lvert \frac{\partial \varphi_{\epsilon}}{\partial x^i}(x-y) - c \pm \right\rvert $. Then

\begin{align} \left\lvert u(y) \left\{ \frac{\varphi_{\epsilon}(x-y+he_i) - \varphi_{\epsilon}(x-y)}{h} \right\} \right\rvert \leq M \: \Bigr\lvert \: u\bigr\vert_{supp\left( \varphi_{\epsilon}(x+\delta e_i - \: \cdot) \right) \cap supp\left( \varphi_{\epsilon}(x- \: \cdot) \right)}(y) \Bigr\rvert . \end{align}

Hence the integrand is bounded by an integrable function.

This is how my question originated.

Answer 1

Yes it is bounded by some constant depending on $\varphi$ itself. Since the support of $\varphi$ is compact you have $$D\varphi(x)=0,\quad x\in \mathbb{R}^n\setminus B_R(0)$$ for a big $R>0$. Since the derivative is also continuous we also have $$\max_{x\in\overline{B_R(0)}} |D\varphi(x)|$$ exists and is therefore finite. Outside of the ball the function and the derivative is zero, hence $$\sup |D\varphi|=\max_{x\in\overline{B_R(0)}} |D\varphi(x)|<\infty.$$