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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.

Regard the space of continuous functions vanishing at infinity $C_0({\mathbb{R}}^n)$ as a subspace of $L^{\infty }({\mathbb{R}}^n)$. Point evaluation at the origin,is a bounded linear functional of norm one on this subspace because a continuous function's supremum and essential supremum agree. By the Hahn-Banach theorem, ${\ell }_0$ extends to a bounded linear functional

Suppose that some $g\in L^1({\mathbb{R}}^n)$ represented this extension:Choose $\phi \in C_c({\mathbb{R}}^n)$ with $0\le \phi \le 1$, $\phi (0)=1$, and support in the unit ball, and set ${\phi }_k(x)=\phi (kx)$. Then $\ell ({\phi }_k)=1$ for every $k$. On the other hand, ${\phi }_k(x)\to 0$ for almost every $x$ and $|{\phi }_{k}g|\le |g|$. The dominated convergence theorem givesa contradiction. Thus $\ell$ is a singular functional on L infinity and has no $L^1$ density.

The Hahn-Banach theorem says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to \mathbb{R}$ is bounded and linear, then there is a bounded linear extension $G:X\to \mathbb{R}$ with $\Vert G\Vert =\Vert g\Vert$.

For $x\in X$, the map $\hat{x}(f)=f(x)$ is linear on $X^{\prime }$ andso $\hat{x}\in X^{\prime \prime }$ and $\Vert \hat{x}\Vert \le \Vert x\Vert$. The assignment $Jx=\hat{x}$ is visibly linear. If $x\ne 0$, define $g$ on $\mathrm{span}\{x\}$ byIt has norm one, and Hahn--Banach extends it to an $f\in X^{\prime }$ of norm one with $f(x)=\Vert x\Vert$. ThereforeThus the canonical embedding into the bidual is an isometry and

On $C([0,1])$, point evaluation $\ell (f)=f(0)$ has norm one for the $L^{\infty }$ norm, because a continuous function's essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda \in (L^{\infty }{)}^{\prime }$. If $\Lambda$ were represented by some $g\in L^1$, thenfor every continuous $f$. Choose continuous $0\le f_n\le 1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the integral makes the right side tend to zero, while the left side is always one, a contradiction. This singular functional on L infinity proves