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$H_0^1$ is functions of $H^1$ that have a property of vanishing (i.e. go to zero) on the boundary of the function's domain.

So $$H_0^1 \subset H^1$$

However, if one considers the sum:

$$v_1 + v_2$$

where $v_1 \in H^1$ and $v_2 \in H_0^1$, then what can one say about the set where this sum is?

I interpret that one cannot say $v_1+v_2 \in H^1$, because there could be more sets in $H^1$. Thus it's possible that the sum would (term-wise) be in $(H^1 -H_0^1) + H_0^1$. Or does this mean the same as in $H^1$?

mavavilj
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    Why can we not say that $v_1+v_2\in H^1$ if $v_1,v_2\in H^1$? –  Jul 12 '18 at 09:03
  • @RyanGibara As I wrote, because I was wondering, whether that's a stretch. If $v_2$ belongs to a subset of $H^1$, then it might not have all the properties of a "full" $H^1$ function. Therefore while it is in $H^1$, then it might not have all the properties of $H^1$? Unless Sobolev spaces guarantee this for subsets? Functions that would have the additive property of linearity would be subsets of linear functions, but they would not be linear, because they don't follow scalar multiplication. – mavavilj Jul 12 '18 at 09:08
  • I think you have the idea backwards. Since $H_0^1\subset H_1$, this means that every element in $H^1_0$ is in $H^1$. This means that functions in $H^1_0$ have all the properties of $H^1$ functions and more. In this case, the more is the vanishing property. –  Jul 12 '18 at 09:11
  • @RyanGibara So this is a property of Sobolev spaces? Because in the linear function case, it's not like that. There's a subset, whose functions don't have the full properties? – mavavilj Jul 12 '18 at 09:11
  • @mavavilj can you give an example where it is not like that (maybe include as an edit in your question)? – supinf Jul 12 '18 at 09:20
  • @mavavilj: This is a property of all vector spaces (or, more generally, additive groups) called closure under addition. I don't think the semi-linear transformations are a subset of linear transformations - I think that the linear transformations are a subset of the semi-linear transformations. –  Jul 12 '18 at 09:22
  • @RyanGibara What's the propertly then in semi-linear that prohibits that if one takes a very large scalar and multiplies some $u+v$ with it, that it doesn't "go out" the semi-linear set? Or if one considers that semi-linear don't even have the scalar multiplication defined, then that would mean that linear functions have a greater number of properties and it'd be odd to consider them a subset of "more limited functions". – mavavilj Jul 12 '18 at 09:26

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From $H_0^1\subset H^1$ and $v_2\in H_0^1$ it follows that $v_2\in H^1$.

Because $H^1$ is a vector space, it means that the sum of elements in $H^1$ is again in $H^1$. This means, that $v_1+v_2\in H^1$ follows from $v_1\in H^1$ and $v_2\in H^1$.

supinf
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  • So are you essentially saying that given vector space $W$ and some $U \subset W$, then $W+U \subset W$? And the Sobolev properties (I'm not very versed at Sobolevs yet) do not alter this. – mavavilj Jul 12 '18 at 14:56
  • @mavavilj yes. Sobolev spaces are vector spaces. – supinf Jul 12 '18 at 15:20