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I am reading a paper regarding the pricing of certain financial derivatives making use of the finite element method. In this paper the following weighted Sobolev spaces are introduced:

$W_{0}$ = $ \{ \phi : \int_{\Omega} \phi^2 dx + \int_{\Omega} x^2 (\phi')^2 dx < \infty \} $

$W_{2,0}$ = $ \{ \phi : \int_{\Omega} \phi^2 dx + \int_{\Omega} x^2 (\phi')^2 \int_{\Omega} x^2 (\phi'')^2 dx < \infty \} $

With associated norms $\Vert . \Vert_{0} $ and $\Vert . \Vert_{2,0} $ , $\Omega = (0,a) $ and where the derivatives are take in the weak sense.

The author then proceeds to give the following result:

$\Vert \phi - \pi_{h}\phi \Vert_{0} \leq ch \Vert \phi \Vert_{2,0} $ $\forall \phi \in W_{2,0}$, where $\pi_{h}$ is the piecewise linear interpolant of $\phi$ over [0,a].

My problem is the definition of $\pi_{h}\phi$ as the functions in the wighted space need not even be defined at 0. So far I have managed to show that the functions in $W_{2,0}$ can be approximated by bounded $C^{\infty}$ functions, are bounded and have a classically differentiable version.

I have seen two possible options:

1) Show the embedding $W_{2,0} \hookrightarrow C(\bar{\Omega})$.

2) Show that $\pi_{h}$ is a bounded linear operator and can thus be extended from the bounded $C^{\infty}$ functions to $W_{2,0}$.

However I can't seem to get either of these two methods to work. Any ideas would be appreciated.

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  • $\begingroup$ If you can prove that functions in $W_{2,0}$ are bounded and can be approximated by $C^\infty$ functions, then the same proof is very likely to show that these functions are continuous. $\endgroup$ – Michael Renardy Sep 7 '14 at 13:23
  • $\begingroup$ It is basically just the right continuity at 0 that is needed, but I couldn't see any way that this followed from boundedness or the density of $C^{\infty}$. $\endgroup$ – User76765 Sep 7 '14 at 14:08
  • $\begingroup$ If the embedding is untrue, it requires a bounded oscillating function, from which it follows that the derivative will be unbounded and oscillate between $+- \infty$ as x approaches 0, similarly for the second derivative. $\endgroup$ – User76765 Sep 7 '14 at 14:24
  • $\begingroup$ Okay, so I have shown that the interpolant is not bounded, so the embedding seems to be the only option. $\endgroup$ – User76765 Sep 9 '14 at 8:17
  • $\begingroup$ How did you prove functions in $W_{2,0}$ are bounded? Did you not do this by showing that the $L^\infty$ norm is bounded in terms of the $W_{2,0}$ norm? You also say you know $C^\infty$ functions are dense. So you have a uniformly convergent sequence of $C^\infty$ functions. This implies the limit is continuous. $\endgroup$ – Michael Renardy Sep 9 '14 at 10:31

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