Let (Omega ) be a bounded domain in (mathbb{R}^d). (H^{-1}(Omega )) is defined as the dual space of (H_0^1(Omega )), which is the space of all bounded linear functionals on (H_0^1(Omega )). Then the (L_2(Omega )) space is a subspace of (H^{-1}(Omega )). This can be verified as below.
Let (f in L_2(Omega )). Define a linear functional using the inner product on (L_2(Omega )) as (f(v) = (f, v)) for all (v in L_2(Omega )). If (v in H_0^1(Omega )), (v) also belongs to (L_2(Omega )). Therefore, the linear functional (f) can also be applied to (v). Because [ orm{v}_1^2 = ( orm{v}_{L_2}^2 + orm{ abla v}_{L_2}^2) geq orm{v}_{L_2}^2, ] [ orm{f}_{-1} = sup _{v in H_0^1(Omega ), v eq 0} frac{abs{f(v)}}{ orm{v}_1} leq sup _{v in H_0^1(Omega ), v eq 0} frac{abs{f(v)}}{ orm{v}_{L_2}} leq sup _{v in L_2(Omega ), v eq 0} frac{abs{f(v)}}{ orm{v}_{L_2}} = orm{f}_{L_2}. ] Therefore, if (f in L_2(Omega )), its (H^{-1}) norm is also finite, hence (f in H^{-1}(Omega )) and (L_2(Omega ) subset H^{-1}(Omega )).
To show (H^{-1}(Omega ) subset L_2(Omega )), an example function can be provided, which belongs to (H^{-1}(Omega )) but not to (L_2(Omega )).
Let (Omega = (0,1)) and (f(x) = frac{1}{x}). (f(x)) has a singularity at (x) and the integral (int _{Omega } f^2 intd x = int _0^1 frac{1}{x^2} intd x) is divergent. Therefore, (f otin L_2(Omega )).
Next, define (f(v) = (f, v)) for any (v in H_0^1(Omega )), we have egin{equation*} egin{aligned} (f, v) &= int _0^1 frac{1}{x} v(x) intd x = int _0^1 v(x) intd (log x) \ &= v(x)log x ig vert _0^1 - int _0^1 log x frac{diff v}{diff x} intd x. end{aligned} end{equation*} Because (v in H_0^1(Omega )), the boundary term in the above integral disappears and [ (f, v) = -int _0^1 log x frac{diff v}{diff x} intd x. ]
Then we want to apply Cauchy-Schwartz inequality to this formula. But before that, we need to verify that both (log x) and (frac{diff v}{diff x}) belong to (L_2(Omega )). Because (v in H_0^1(Omega )), the latter belongs to (L_2(Omega )) for sure. For the former, we need to show (int _0^1 (log x)^2 intd x) is finite.
We notice that [ frac{diff (xlog x - x)}{diff x} = log x + x frac{1}{x} - 1 = log x ] and [ frac{diff (xlog ^2 x)}{diff x} = log ^2 x + x (2log x) frac{1}{x} = log ^2 x + 2log x. ] Then we have [ frac{diff (xlog ^2 x - 2xlog x + 2x)}{diff x} = log ^2 x. ] Hence, using l’Hospital’s rule, [ int _0^1 log ^2 x intd x = 2 ] and (log x in L_2(Omega )). Now we apply Cauchy-Schwartz inequality, [ abs{(f, v)} = igabs{int _0^1 log x frac{diff v}{diff x} intd x} leq orm{log x}_{L_2} ignorm{frac{diff v}{diff x}}_{L_2} = sqrt{2} abs{v}_1. ] Therefore, (sup frac{abs{f(v)}}{abs{v}_1} leq sqrt{2}) and (f in H^{-1}(Omega )). (L_2(Omega )) is a proper subspace of (H^{-1}(Omega )).
References