Approximation of $u \in C^{1,2}([0,T]\times \mathbb{R}^d)$ with $u(0,x)= 0$ by smooth functions vanishing close to $t=0$

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!

• 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. Jul 3, 2020 at 23:01
• 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? Jul 5, 2020 at 11:18

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.

• 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 $[0,T]$. Jul 8, 2020 at 15:16