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The Sobolev embedding theorem as stated in my notes says that if we have $k > l + d/2$ then we can continuously extend the inclusion $C^\infty(\mathbb T^d) \hookrightarrow C^l(\mathbb T^d)$ to $H^k(\mathbb T^d) \hookrightarrow C^l(\mathbb T^d)$.

We define $H^k$ to be the closure of $C^\infty$ with respect to the Sobolev norm, see my previous question for the definition.

What I'm confused about is, why we need the condition $k > l + d/2$. What exactly does it give us? If $H^k$ is the closure of $C^\infty$ we already get that if $T$ is any continuous linear operator $C^\infty \to C^l$ we can extend it to all of $H^k$. What am I missing? Thanks for your help.

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You said that if $T$ is any continuous linear operator $C^\infty \to C^l$ we can extend it to all of $H^k$. This is true, provided that the topology on $C^\infty$ is inherited from that of $H^k$. In other words, one has to use $H^k$-norms in the domain to define continuity of $T:C^\infty\to C^l$. The condition $k>\frac{n}2+l$ guarantees this continuity when $T$ is the canonical inclusion.

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    Thank you! I think I have to think of $i: C^\infty \hookrightarrow C^l$ as a map taking $(f,0, \dots, 0)$ to $(f, 0, \dots, 0)$ then it's clear that the norm on $C^\infty$ has to be the Sobolev norm. The other map $C^\infty \to C^\infty$ mapping $f$ to $(f, 0 , \dots, 0)$ is also an inclusion but a different one. – 2012-08-18
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Suppose that the linear embedding $(C^\infty(\mathbb T^d),\|\,\cdot\,\|_{H^k}) \hookrightarrow (C^l(\mathbb T^d),\|\,\cdot\,\|_{C^l})$ is continuous from the incomplete space on the left to the complete space on the right.

Then $i$ lifts to a unique, continuous linear map $\bar i: (H^k,\|\,\cdot\,\|_{H^k}) \rightarrow (C^l(\mathbb T^d),\|\,\cdot\,\|_{C^l})$, where $(H^k,\|\,\cdot\,\|_{H^k})$ is the completion of $(C^\infty(\mathbb T^d),\|\,\cdot\,\|_{H^k})$.

To get an embedding theorem, we need to check that $\bar i$ is injective. This property is not automatic, and fails in general.

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    @NateEldredge I assumed that the OP had chosen parameters so that the inclusion $i$ was already continuous when $C^\infty$ is equipped with the $H^k$ norm. Maybe I misunderstood the question. – 2012-08-15
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The elements which are in $H^k∖C^{\infty}$ could be mapped by the extension to a function which may not be $C^l$. Sobolev embedding say that it's actually the case with a condition on the dimension and the order of derivatives we want to work with.

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    That doesn't quite make sense: the extension, if it exists, will by definition be a map from $H^k$ to $C^l$. – 2012-08-15