Inner product and norm (Q2732196)

Some equivalent forms of the norm in inner product spaces are proved. The results proved are:NEWLINENEWLINENEWLINE(i) A normed linear space \(X\) is an inner product space if and only if any one of (1) to (4) holds:NEWLINENEWLINENEWLINE(1) \(\|x_1- x_2\|^2\leq\|x_1\|^2+\|x_2\|^2\leq\|x_1+ x_2\|^2\) or \(\|x_1+ x_2\|^2\leq\|x_1\|^2+\|x_2\|^2\leq\|x_1- x_2\|^2\), \(\forall x_1,x_2\in X\),NEWLINENEWLINENEWLINE(2) \(\sum_{e^n} \|\sum^n_1 e_i x_i\|^2= 2^n \sum^n_1\|x_i\|^2\), \(\forall x_i\in X\), \(e^n=(e_1,e_2,\dots, e_n)\), \(e_i=\pm 1\), \(1\leq i\leq n\), \(n\geq 2\),NEWLINENEWLINENEWLINE(3) \(\inf_{e^n} \|\sum^n_1 e_ix_i\|^2\leq \sum^n_1\|x_i\|^2\leq \sup_{e^n} \|\sum^n_1 e_ix_i\|^2\),NEWLINENEWLINENEWLINE(4) there exists a function \(f(t)\), \(t\geq 0\), \(f(0)= 0\), \(f(1)= 1\) such that \(\sum_{e^n} f(\|\sum^n_1 e_ix_i\|)= 2^n \sum^n_1 f(\|x_i\|)\), \(1\leq i\leq n\).NEWLINENEWLINENEWLINE(ii) A normed linear space is a real inner product space if and only if any one of (5) or (6) holds:NEWLINENEWLINENEWLINE(5) \((\sum^n_1 a^2_i) (\sum^n_1\|x_i\|^2)= \|\sum^n_1 a_ix_i\|^2+ \sum_{1\leq i\leq n}\|a_i x_j- a_j x_i\|^2\) for any real number \(a_1,a_2,\dots, a_n\) and \(x_i\in X\), \(1\leq i\leq n\), norNEWLINENEWLINENEWLINE(6) there exists a continuous function \(f(t)\) for \(t\geq 0\), \(f(0)= 0\), \(f(1)= 1\) such that NEWLINE\[NEWLINE\Biggl(\sum^n_1 a^2_i\Biggr) \Biggl(\sum^n_1 f(x_i)\Biggr)= f\Bigl(\bigl\|\sum^n_1 a_i x_i\bigl\|\Bigr)+ \sum_{1\leq i\leq n} f(\|a_i x_j- a_jx_i\|).NEWLINE\]