Why Are Higher-Order Derivatives Needed for Continuity in Sobolev Spaces?

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Hello there,

I have a question about Sobolev's first theorem.

Essentially it is said that for p = 1 the space [tax] W^{l}_{p} [\tex] belongs to the space of continuous if l ≥ n.
So, in 1D I just need the finction and its first derivative to be integrable, which makes perfect sense.
I can not get why though for n > 1 I need information on derivatives of higher order to conclude the finction is continuous. In other words, can anyone show me a counterexample of a function [tex]W^{1}_{1} [\tex] not continuous over a 2D domain?<br /> <br /> The second curiosity concerns the following statement I have found on the web, <a href="http://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/</a><br /> "<br /> (One can already see this with the most basic example of Sobolev embedding, coming from the fundamental theorem of calculus. If a (continuously differentiable) function f has f' in L1 then we of course have f belongs to L∞ ; but the converse is far from true.) "<br /> <br /> Now the Sobolev theroem tell us that in order for a function to be L∞, in 1D, we just need it to belong to [tex]W^{1}_{1} [\tex], which seems to me larger then the class of continuoulsy differentiable function, am I missing something?[/tex][/tex]
 
on Phys.org
Hi there,

I am starting to wonder if my question was unclear, or meaningless..
If so please let me know and I will reformulate it.

Many thanks
 

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