My question is the following: Let $T>0$, $u: [0,T]\times \mathbb{R}^d \to \mathbb{R}$ be continuously differentiable once w.r.t. $t \in (0,T)$ and twice w.r.t. $x \in \mathbb{R}^d$ such that all partial derivatives can be extended to $[0,T]\times \mathbb{R}^d$ continuously. Further, suppose $u$ along with all its partial derivatives is bounded.

Does there exist a sequence $(u_n)_n$ with $u_n \in C^{\infty}_c((0,T]\times \mathbb{R}^d)$ which approximates $u$ pointwise up to all $(1,2)$-derivatives such that each sequence of derivatives is bounded in $n$?

The answer is affirmative, if we do not require $u_n$ to vanish close to $0$ (note that there is no assumption on vanishing close to $T$), because clearly $u$ as above can be approximated by smooth functions in the desired sense. However, to me it seems possibly troublesome to prove what I'm looking for, essentially because if we require $\partial_tu_n$to be bounded uniformly in $n$, we cannot allow each $u_n$ to be constantly $0$ on some $(\epsilon_n,T]\times \mathbb{R}^d$, because this might require us to approximate $u$ by functions $u_n$ with increasing (not bounded) derivative $\partial_tu_n$ close to $t=\epsilon_n$.

Is it still possible to obtain what I'm looking for? My hope is that due to $u(0,x)=0 \forall x$ the situation is actually not too bad. Any help is appreciated!

  • $\begingroup$ For any continuous function $u$, the convolution of mollifier and the $u$ is smooth and converges to $u$ uniformly on any bonded domain in $\mathbb{R}^d$ . So, it is point-wise. $\endgroup$ Jul 3, 2020 at 23:01
  • $\begingroup$ Sure. However, such a mollification is in general not compactly supported in $(0,T]\times \mathbb{R}^d$. I guess this is precisely what makes the answer to my question non-trivial or am I mistaken? $\endgroup$ Jul 5, 2020 at 11:18

1 Answer 1


Let $x_a,a=1,2,3,..$ be points of $\mathbb{R}^d$ such that Balls of raius 1 is an locally finite open covering: $\mathbb{R}^d =\cup \mathbb{B}(x_a,1)$, where by the term locally finite, we mean that for any $x \in \mathbb{R}^d$, the cardinality of the set $A(x):= \{a; x \in \mathbb{B}(x_a,1) \}$ is finite.

Let $\rho_a$ be the partition of unity for the covering, i,e,, we assume the following properties:

  1. $\Sigma \rho_a(x) =1 $ for all $x \in \mathbb{R}^d$.
  2. supp $\rho_a \subset \mathbb{B}(x_a,1)$.

Note that the following lemma.

Let $\phi(x) = c\exp(\frac{1}{|x|^1-1})$ for$|x| \leq 1$ and $0$ for $|x| \geq 1$ and $c$ is chosen so that $\int _{\mathbb{R} ^d}\phi(x) > dx=1$. Let $\mathbb{\Omega}$ be a dommain in $\mathbb{R}^d$. Define $u_h:= \frac{1}{h^{-d}} \int_\Omega \phi(\frac{x-y}{h})u(y)dy$ for any $h < \text{dist}(x, \partial \Omega)$. Then $u_h$ converges to $u$ uniformly on any domain $\Omega ' \subset \subset \Omega$. Note that $u_h \in C^\infty(\Omega ')$.

We use this lemma on each ball $ \mathbb{B}(x_a,1)$.

Let $v \in C^0(\mathbb{R}^d)$ and let $v_h^a:= \frac{1}{h^{-d}} \int_{\mathbb{B}(x_a,2) }\phi(\frac{x-y}{h})u(y)dy$. Then, The above lemma implies that $v_h^a$ converges to $v$ uniformly on $\mathbb{B}(x_a,1)$.

Define $v_h:= \Sigma \rho_a v_h^a$, then for any $x \in \mathbb{R}^d$,

$$|v_h(x)-v(x)| \leq \Sigma_{a \in A(x)}\rho_a(x) |v_h^a(x)-v(x)| $$ Because the set $A(x)$ is finite, there is some $a^*$ so that

$$ \leq (\#A(x))|v_h^{a^*}(x)-v(x)| $$

$$ \leq \#A(x)\|v_h^{a^*}-v\|_{L^{\infty}(\mathbb{B}(x_{a^*},1))}.$$

So, the function $v_h$ coverges to $v$ in pointwise sense.

Because $u \in C^{1,2}([0,T]\times \mathbb{R}^d)$ is also $u \in C^{0,0}([0,T]\times \mathbb{R}^d)$, we can apply the above for it.

  • $\begingroup$ The approximating sequence you considered does not necessarily fulfill the uniform boundedness-condition of the derivatives which I require. This is easily fixed (even without consideration of partition of unity) in $\mathbb{R}^d$ but may cause problems for the closed interval $[0,T]$. $\endgroup$ Jul 8, 2020 at 15:16

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