我们这里只给出实赋范向量空间的情况的证明； 复赋范向量空间的情况的证明是类似的。
通过上述公式，如果范数由内积描述（如我们所希望的那样），那么它必须满足
\[\langle x, \ y \rangle = \frac{1}{4} \left(\|x+y\|^2 - \|x-y\|^2\right) \quad \text{ for all } x, y \in H.\]
还有待证明这个公式定义了一个内积，并且这个内积导出范数 \(\|\cdot\|.\)
明确地讲，要证明以下内容：
\begin{array}l
\text{1.}\langle x, x \rangle = \|x\|^2, \quad x \in H\\
\text{2.}\langle x, y \rangle = \langle y, x \rangle, \quad x, y \in H\\
\text{3.}\langle x+z, y\rangle = \langle x, y\rangle + \langle z, y\rangle \quad \text{ for all } x, y, z \in H,\\
\text{4.}\langle \alpha x, y \rangle = \alpha\langle x, y \rangle \quad \text{ for all } x, y \in H \text{ and all } \alpha \in \R
\end{array}
(This axiomatization omits positivity, which is implied by (1) and the fact that \(\|\cdot\|\) is a norm.)
For properties (1) and (2), substitute: \(\langle x, x \rangle = \frac{1}{4} \left(\|x+x\|^2 - \|x-x\|^2\right) = \|x\|^2,\) and \(\|x-y\|^2 = \|y-x\|^2.\)
For property (3), it is convenient to work in reverse.
It remains to show that
\[\|x+z+y\|^2 - \|x+z-y\|^2 \overset{?}{=} \|x+y\|^2 - \|x-y\|^2 + \|z+y\|^2 - \|z-y\|^2\]
or equivalently,
\[2\left(\|x+z+y\|^2 + \|x-y\|^2\right) - 2\left(\|x+z-y\|^2 + \|x+y\|^2\right) \overset{?}{=} 2\|z+y\|^2 - 2\|z-y\|^2.\]
现在应用平行四边形恒等式：
\[2\|x+z+y\|^2 + 2\|x-y\|^2 = \|2x+z\|^2 + \|2y+z\|^2\]
\[2\|x+z-y\|^2 + 2\|x+y\|^2 = \|2x+z\|^2 + \|z-2y\|^2\]
因此还有待证明：
\[\cancel{\|2x+z\|^2} + \|2y+z\|^2 - (\cancel{\|2x+z\|^2} + \|z-2y\|^2) \overset{?}{{}={}} 2\|z+y\|^2 - 2\|z-y\|^2\]
\[\|2y+z\|^2 - \|z-2y\|^2 \overset{?}{=} 2\|z+y\|^2 - 2\|z-y\|^2\]
这个等式可以通过以下两个平行四边形恒等式相减来证明：
\[\|2y+z\|^2 + \|z\|^2 = 2\|z+y\|^2 + 2\|y\|^2\]
\[\|z-2y\|^2 + \|z\|^2 = 2\|z-y\|^2 + 2\|y\|^2\]
Thus (3) holds.
It can be verified by induction that (3) implies (4), as long as \(\alpha \in \Z.\)
But "(4) when \(\alpha \in \Z\)" implies "(4) when \(\alpha \in \Q\)".
And any positive-definite, real-valued, \(\Q\)-bilinear form satisfies the Cauchy–Schwarz inequality,
so that \(\langle \sdot,\sdot \rangle\) is continuous.
Thus \(\langle \sdot,\sdot \rangle\) must be \(\R\)-linear as well.
