The Analytic Hahn Banach Theorem - Examples

Apply [L1] to the nonzero vector $f\in L^p(\mu )$. This gives a functional $\mathrm{\Phi }\in (L^p(\mu ){)}^{\ast }$ with $\mathrm{\parallel }\mathrm{\Phi }\mathrm{\parallel }=1$ and $\mathrm{\Phi }(f)=\mathrm{\parallel }f{\mathrm{\parallel }}_p$.

By [L2], the same number $\mathrm{\parallel }f{\mathrm{\parallel }}_p$ is obtained as $\sup \{\mid \int fgd\mu \mid :\mathrm{\parallel }g{\mathrm{\parallel }}_q\le 1\}\mathrm{.}$ In the standard explicit realization, one takes $g=\mathrm{\mid }f{\mathrm{\mid }}^{p-1}\mathrm{sgn}f\mathrm{/}\mathrm{\parallel }f{\mathrm{\parallel }}_p^{p-1}$ when $1<p<\mathrm{\infty }$, and $g=\mathrm{sgn}f$ when $p=1$.

Thus the abstract existence statement from Hahn-Banach and the concrete $L^p-L^q$ formula from the earlier page identify the same norming phenomenon: one proves that some unit functional attains $\mathrm{\parallel }f{\mathrm{\parallel }}_p$, and the other writes an attaining functional down explicitly.

For $(t,t)\in M$ one has $\mathrm{\mid }{f}_0(t,t)\mathrm{\mid }=\mathrm{\mid }t\mathrm{\mid }=\mathrm{\parallel }(t,t){\mathrm{\parallel }}_{\mathrm{\infty }},$ so $\mathrm{\parallel }{f}_0\mathrm{\parallel }=1$. Also every $(a,b)\in {\mathbb{R}}^2$ decomposes as $(a,b)=\frac{a+b}{2}(1,1)+\frac{a-b}{2}(1,-1),$ so $X=M\oplus \mathbb{R}(1,-1)$.

If $F$ is a linear extension of $f_0$ and $c:=F(1,-1)$, then step 1.1 forces $F(a,b)=\frac{a+b}{2}F(1,1)+\frac{a-b}{2}F(1,-1)=\frac{1+c}{2}a+\frac{1-c}{2}b\mathrm{.}$ Conversely, the displayed formula defines a linear functional extending $f_0$.

Let $\phi (a,b):=\alpha a+\beta b$ on $({\mathbb{R}}^2,\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }})$. If $\mathrm{\parallel }(a,b){\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, then $\mathrm{\mid }\phi (a,b)\mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }\mathrm{.}$ Choosing $a=\mathrm{sgn}\alpha$ and $b=\mathrm{sgn}\beta$ (interpreting $\mathrm{sgn}0=0$) gives $\mathrm{\parallel }(a,b){\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{\mid }\phi (a,b)\mathrm{\mid }=\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }$, so $\mathrm{\parallel }\phi \mathrm{\parallel }=\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }$. Applying this to the formula from step 2.1 yields $\mathrm{\parallel }{F}_c\mathrm{\parallel }=\frac{\mathrm{\mid }1+c\mathrm{\mid }+\mathrm{\mid }1-c\mathrm{\mid }}{2}\mathrm{.}$ This equals $1$ exactly when $-1\le c\le 1$.

Therefore every $c\in [-1,1]$ gives a norm-preserving extension of $f_0$. This computes explicitly the admissible interval predicted abstractly by [L1], and in particular is consistent with the existence statement of [L2].

Define $p:{\mathrm{\ell }}^{\mathrm{\infty }}\to \mathbb{R}$ by $p(x):={\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{\sigma }_n(x)\mathrm{.}$ If $x\in {\mathrm{\ell }}^{\mathrm{\infty }}$, choose $M\ge 0$ with $\mathrm{\mid }{x}_n\mathrm{\mid }\le M$ for all $n$; then every Cesaro mean satisfies $\mathrm{\mid }{\sigma }_n(x)\mathrm{\mid }\le M$, so [L4] shows that $p(x)$ is an ordinary real number. Because ${\sigma }_n(x+y)={\sigma }_n(x)+{\sigma }_n(y)$ for every $n$, [L5] gives $p(x+y)\le p(x)+p(y)\mathrm{.}$ Also ${\sigma }_n(tx)=t{\sigma }_n(x)$ for every $t\ge 0$, so $p(tx)=tp(x)$. Thus $p$ is sublinear in the sense of [L1].

Let $C\subseteq {\mathrm{\ell }}^{\mathrm{\infty }}$ be the set of sequences whose Cesaro means converge, and define $\lambda (x):={\lim }_{n\to \mathrm{\infty }}{\sigma }_n(x)(x\in C)\mathrm{.}$ Since ${\sigma }_n(ax+by)=a{\sigma }_n(x)+b{\sigma }_n(y)$ for all real scalars $a,b$, the set $C$ is a linear subspace and $\lambda$ is linear. If $x\in C$, then the convergent sequence $({\sigma }_n(x))$ has limit superior equal to its limit, so $\lambda (x)\le p(x)$. Therefore [L2] yields a linear extension $L:{\mathrm{\ell }}^{\mathrm{\infty }}\to \mathbb{R}$ of $\lambda$ with $L\le p$ on all of ${\mathrm{\ell }}^{\mathrm{\infty }}$.

Let $x=(x_n)\in {\mathrm{\ell }}^{\mathrm{\infty }}$. Since $x$ is bounded, there is $M\ge 0$ with $\mathrm{\mid }{x}_n\mathrm{\mid }\le M$ for all $n$. Using [L3], ${\sigma }_n(x-Sx)=\frac{1}{n+1}{\sum }_{k=0}^n(x_k-x_{k+1})=\frac{x_0-x_{n+1}}{n+1}\mathrm{.}$ Hence $\mid {\sigma }_n(x-Sx)\mid \le \frac{\mathrm{\mid }{x}_0\mathrm{\mid }+M}{n+1}\to 0,$ so $x-Sx\in C$ and $\lambda (x-Sx)=0$. Since $L$ extends $\lambda$, $L(x-Sx)=0$, that is, $L(Sx)=L(x)$.

Step 2.1 gives the extension and domination properties, and step 3.1 gives shift invariance. Therefore $L$ has all three properties listed in the example.

Shift invariance is part of [L1]. Let $\mathbf{1}=(1,1,1,\dots )$. By [L3], every Cesaro mean of $\mathbf{1}$ equals $1$, so the extension property in [L1] gives $L(\mathbf{1})=1$. By linearity, $L(c\mathbf{1})=c(c\in \mathbb{R})\mathrm{.}$

Suppose $x_n\ge 0$ for every $n$. Then every Cesaro mean of $-x$ is nonpositive, so ${\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{\sigma }_n(-x)\le 0.$ The domination part of [L1] therefore gives $-L(x)=L(-x)\le {\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{\sigma }_n(-x)\le 0,$ hence $L(x)\ge 0$. So $L$ is positive.

Let $M:=\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$. Then $-M\mathbf{1}\le x\le M\mathbf{1}$ termwise, so the sequences $M\mathbf{1}-x$ and $M\mathbf{1}+x$ are pointwise nonnegative. By step 1.2, $0\le L(M\mathbf{1}-x)=M-L(x)\text{and}0\le L(M\mathbf{1}+x)=M+L(x)\mathrm{.}$ Thus $\mathrm{\mid }L(x)\mathrm{\mid }\le M=\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $\mathrm{\parallel }L\mathrm{\parallel }\le 1$. Since step 1.1 gives $L(\mathbf{1})=1$ and $\mathrm{\parallel }\mathbf{1}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, one also has $\mathrm{\parallel }L\mathrm{\parallel }\ge 1$. Therefore $\mathrm{\parallel }L\mathrm{\parallel }=1$.

Write $\alpha :={\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{x}_n$ and $\beta :={\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{x}_n$. Let $\epsilon >0$. By the definition in [L2], there is $N\in \mathbb{N}$ such that for all $k\ge N$, $\alpha -\epsilon \le x_k\le \beta +\epsilon \mathrm{.}$ Hence every term of the shifted sequence $S^{N}x$ lies between the constant sequences $(\alpha -\epsilon )\mathbf{1}$ and $(\beta +\epsilon )\mathbf{1}$. Using positivity from step 1.2, the constant-sequence values from step 1.1, and shift invariance from [L1], we get $\alpha -\epsilon =L((\alpha -\epsilon )\mathbf{1})\le L(S^{N}x)=L(x)\le L((\beta +\epsilon )\mathbf{1})=\beta +\epsilon \mathrm{.}$

Since the inequalities of step 2.2 hold for every $\epsilon >0$, one obtains $\alpha \le L(x)\le \beta$. Together with steps 1.1, 1.2, and 2.1, this shows that $L$ is a Banach limit.

By [L1], the functionals $F_1(a,b)=a\text{and}{F}_{-1}(a,b)=b$ are both norm-preserving extensions of $f_0$.

They agree on $M$, since $F_1(t,t)=t=F_{-1}(t,t)$ for every $(t,t)\in M$, but they differ on $(1,-1)$: $F_1(1,-1)=1\mathrm{\ne }-1=F_{-1}(1,-1)\mathrm{.}$

Thus the same functional on the same subspace has two distinct norm-preserving Hahn-Banach extensions. Therefore the uniqueness claim is false.