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The Lp norm of the forcing at the critical exponent is constant:To prove weak convergence in $L^{6/5}(B)$, the duality of Lp spaces says that we must test against an arbitrary $g\in L^6(B)$. Holder inequality, restricted to the shrinking measurable set $E_m$, givesThe last step is absolute continuity of the Lebesgue integral, since $|E_m|=r_m^3|E|\to 0$ and $|g{|}^6$ is integrable. ThusTesting only bounded or smooth functions would not by itself establish this weak convergence; the restricted Holder inequality handles every member of the full dual space.

By part i, ${\chi }_{E_n}|g|\to 0$ almost everywhere, while $0\le {\chi }_{E_n}|g|\le |g|$ and $g$ is Lebesgue integrable. The dominated convergence theorem givesTo deduce absolute continuity of the Lebesgue integral, suppose the stated uniform conclusion failed for some ${\epsilon }_0>0$. For each $n$ one could choose a measurable set $A_n$ with $m(A_n)<2^{-n}$ but ${\int }_{A_n}|g|\ge {\epsilon }_0$. Applying the result just proved to this sequence contradicts the lower bound. Hencefor every measurable set $A$.

On $C_c^0({\mathbb{R}}^n)$ equipped with the $L^{\infty }$ norm, point evaluation is linear and satisfieswith equality for a compactly supported function equal to one near zero. Thus $\Vert \Lambda \Vert =1$. The Hahn-Banach theorem extends it, without increasing its norm, to a continuous functional $\tilde{\Lambda }$ on $L^{\infty }({\mathbb{R}}^n)$.

This extension cannot be represented by an $L^1$ function. If $h\in L^1$ satisfied $\tilde{\Lambda }(f)=\int fh$, choose continuous functions $0\le f_k\le 1$ with $f_k(0)=1$ and support in the ball of radius $1/k$. Thenby the absolute continuity of the Lebesgue integral, a contradiction. Therefore $(L^{\infty }{)}^{\prime }$ strictly contains the functionals represented by $L^1$; this is a singular functional on L infinity.

Finally, $L^p({\mathbb{R}}^n)$ is a reflexive Banach space exactly whenThe endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive.