Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does being in C2 imply being in L2 for a function?

  1. Mar 22, 2013 #1
    Having bounded set ##U\subset\mathbb{R}^n## with ##C^1## boundary ##\partial U## and a function ##g\in C^2(\partial U)##, does one automatically have ##g\in L^2(\partial U)##?
    I don't need a proof/explanation, yes/no answer is sufficient.
     
  2. jcsd
  3. Mar 22, 2013 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I assume ##C^2## is twice continuously differentiable and that ##L^2## square (Lebesgue)-integrable.

    The answer is no. An example is given by some modification of the function ##f(x)=\frac{\sin (x)}{x}##.

    If ##\partial U## is compact, then the answer is probably yes.
     
  4. Mar 22, 2013 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Since the domain (U) is bounded, then your example is bad. U bounded and the condition on the boundary may be sufficient to insure continuity on the closure of U, which is compact.
     
  5. Mar 23, 2013 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I somehow missed that ##U## was supposed to be bounded. The result is true then.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook