Properties of a functional on $\ell^\infty$

functional-analysis

It seems you are considering only real sequences.

If you have (a)+(b), take any $x$. Then $(\|x\|\pm x_1,\|x\|\pm x_2,\ldots)\geq0$. This tells us that $$ 0\leq\Phi(\|x\|\pm x_1,\|x\|\pm x_2,\ldots)=\|x\|\pm\Phi(x). $$ Thus $|\Phi(x)|\leq\|x\|$.

For the converse, suppose that $\|x\|\leq1$ and $x\geq0$. So $0\leq x_j\leq 1$ for all $j$. Then $-1\leq 2x_j-1\leq 1$. This shows that $\|2x-1\|\leq1$. Thus $$ |2\Phi(x)-1|=|\phi(2x-1)|\leq1. $$ Unwinding the above, $$ 0\leq\Phi(x)\leq 1, $$ and $\Phi(x)\geq0$. Now for arbitrary $x$ we apply the above to $x/\|x\|$.

With a bit more work, the result also applies to the case of complex sequences.

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