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Consider the $\mathbb R-$linear space $B(X,\mathbb C)$ of all bounded functions from $X\to\mathbb C$ for some $X\ne\emptyset.$ Then $\displaystyle||.||:B(X,\mathbb C)\to\mathbb R:f\mapsto\sup_{x\in X}|f(x)|$ is a norm on $B(X,\mathbb C).$

Is't possible to define a real inner product $\langle,\rangle: B(X,\mathbb C)\times B(X,\mathbb C)\to\mathbb R$ such that it generates the above norm?

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To check if a norm is induced by some sort of inner product, you need to check this identity. If this identity does not hold in general, then it cannot be induced by an inner product. –  Cameron Williams Jul 3 '13 at 17:04
    
Addendum: to show it does not hold, all you need to do is find two such functions that do not obey the equality. –  Cameron Williams Jul 3 '13 at 17:27

1 Answer 1

up vote 1 down vote accepted

If $X$ consists of more than one point then it is not going to happen. Assume $a,b\in X$. Then you can have $f(x)=0$, for $x\neq b$, $g(x)=0$, for $x\neq a$, and $f(b)=g(a)=||f||=||g||=1$. Then

$$||f+g||^2+||f-g||^2=|f(a)+g(a)|^2+|f(a)-g(a)|^2=2||f||^2=2\neq4=2||f||^2+2||g||^2.$$

For $X=\{a\}$ then we have

\begin{align}||f+g||^2+||f-g||^2&=|f(a)+g(a)|^2+|f(a)-g(a)|^2\\&=(f(a)+g(a))(\overline{f(a)}+\overline{g(a)})+(f(a)-g(a))(\overline{f(a)}-\overline{g(a)})\\&=2f(a)\overline{f(a)}+2g(a)\overline{g(a)}\\&=2|f(a)|^2+2|g(a)|^2\\&=2||f||^2+2||g||^2.\end{align}

Therefore it comes from an scalar product.

More directly, it comes from the scalar product $$\left<f,g\right>:=f(a)\overline{g(a)}.$$

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