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For $1\leqslant p\leqslant \infty$, $\Omega$ an open set, define the homogeneous Sobolev space $$\dot{W}^{1,p}(\Omega)=\{u\in L_{loc}^1(\Omega): D^\alpha u\in L^p(\Omega),\ |\alpha|=1\}.$$

Let $\varphi\in C_c^\infty(\Omega)$ which is the smooth function with compact support. My question is, given $f\in \dot{W}^{1,\infty}(\Omega)$, whether we can get $f\varphi\in \dot{W}^{1,\infty}(\Omega)$ or not.

Trump
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1 Answers1

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Yes, this is true. The proof is below.

Beware nevertheless that you cannot have an inequality of the form $$ | f \varphi |_{\dot{W}^{1,\infty}(\Omega)} \leq \| \varphi \|_{C^1} \cdot |f|_{\dot{W}^{1,\infty}(\Omega)} \quad \text{i.e.} \quad \| D^\alpha ( f \varphi ) \|_{L^\infty(\Omega)} \leq \| \varphi \|_{C^1} \| D^\alpha f \|_{L^\infty(\Omega)} $$ because you might have $f(x) = C$ a constant, for which the right-hand sides are $0$, but the left-hand sides might be arbitrarily large.

Now the proof. First, $f \varphi \in L^1_{loc}(\Omega)$ because $f \in L^1_{loc}(\Omega)$ and $\varphi \in L^\infty(\Omega)$. Moreover, from the Leibniz formula: $$ D^\alpha(f\varphi) = (D^\alpha f) \varphi + f (D^\alpha \varphi). $$ The first term is easy: $D^\alpha f \in L^\infty(\Omega)$ was the assumption, and $\varphi \in L^\infty(\Omega)$ also, so their product too. For the second term, since $D^\alpha \varphi \in L^\infty(\Omega)$ and is compactly supported, we only need to check that $f \in L^\infty_{loc}(\Omega)$.

Let $K \subset \Omega$ be a compact set. Up to a finite covering, we can assume that $K$ is a closed ball of radius $r > 0$. We want to prove that $f \in L^\infty(K)$. Taking any pair of points $a,b$ in the ball, we have $$ |f(b)-f(a)| \leq 2r \|D f\|_{L^\infty(K)} $$ hence $$ |f(b)| \leq |f(a)| + 2r \|D f\|_{L^\infty(K)}. $$ Averaging over $a \in K$, we obtain $$ |f(b)| \leq C \| f \|_{L^1(K)} + C\|D f\|_{L^\infty(K)}. $$ Since $f \in \dot{W}^{1,\infty}(\Omega)$, the right-hand side is finite, so $f \in L^\infty(K)$. Hence $f \in L^\infty_{loc}(\Omega)$, which concludes the proof.

cs89
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  • Thanks for your response! But I have two points that I don't really understand. (1) The first is why "up to a finite covering, we can assume that K is a closed ball of radius r>0".
    (2) The second is why we can have "|f(b)-f(a)|≤2r||Df|L∞(K)"? It seems if we don't assume K is a closed ball, we can replace 2r by diam(K) (i.e. the diamter of K). I know this inequality is true if f∈C1(Ω) and without assume K is a closed ball.     – Trump May 12 '23 at 06:26
  • There is some flexibility in this part of the proof depending on your exact assumptions. Your result probably requires that $K$ is convex no? To obtain $|f(a)-f(b)| \leq |a-b| | Df |_{L^\infty}$, you just write $f(a) - f(b) = \int_0^1 (a-b) \cdot \nabla f(b+t(a-b)) \mathrm{d} t$ and use triangular inequality. To prove that $f$ is bounded on $K$ compact, you can cover $K$ by a finite number of balls, then prove that $f$ is bounded on each of these. – cs89 May 12 '23 at 07:53
  • $f(a)-f(b)=\int_0^1(a-b)\cdot\nabla f(b-t(a-b)),\mathrm{d}t$ requires that $f$ is $C^1$, but here we don't know where $f$ is $C^1$. Maybe we can use convolution of $f$ and then approximate $f$ by this convolution? I know the definition of compact, but I don't know exactly why we can assume $K$ is a closed ball. – Trump May 12 '23 at 10:06
  • You don't need $f \in C^1$ to write this. This is a separate question probably already asked multiple times here. 2) Let $K \subset \Omega$ be a compact set. Since $\Omega$ is open, for each $x \in K \subset \Omega$, you can find a small ball $B_x \subset \Omega$ centered on $x$. Then $K \subset \cup_{x \in K} B_x$. Since $K$ is compact, by the open cover definition, there are a finite number of these balls such that $K \subset \cup_{i=1}^n B_{x_i}$. So you only need to prove that $f$ is bounded on each $B_{x_i}$.
    • – cs89 May 12 '23 at 10:17
  • I understand the second point. Could you provide me with some more information or a link for the first point? Textbooks near my hand all assume $f\in C^1$ (or even $C^\infty$) when they use this inequality. Or they may use this inequality for the convolution of $f$. – Trump May 12 '23 at 10:50
  • Using a regularization is a good idea. You can do it at the end. The reasoning above proves that $|f|{L^\infty(B)} \leq |f|{L^1(B)} + |Df|{L^\infty(B)}$ when $f \in C^1$. Now if $f \in L^1$ with $Df \in L^\infty$, consider $f_n = f \star \rho_n$ so that $|f_n|{L^1} \leq |f|{L^1}$ and $|Df_n|{L^\infty}\leq| Df|{L^\infty}$. Then you get a uniform bound for $|f_n|{L^\infty}$ and you know that $f_n(x) \to f(x)$ almost everywhere (see e.g. here), so the same bound holds for $f$. – cs89 May 12 '23 at 11:42