Norm satisfying Parallelogram Law induced by Inner Product/Lemma

Lemma

Let $V$ be a vector space over $\R$.

Let $\norm \cdot : V \to \R$ be a norm on $V$ such that:

$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$

for each $x, y \in V$.

Define the function $\innerprod \cdot \cdot : V \times V \to \R$ by:

$\ds \innerprod x y = \frac {\norm {x + y}^2 - \norm {x - y}^2} 4$

for each $x, y \in V$.

Then:

$\innerprod {n x} y = n \innerprod x y$

for each integer $n \ge 0$ and $x, y \in V$.

Proof

Let $x, y \in V$.

In the case $n = 0$, we have:

$\innerprod {n x} y = \innerprod 0 y$

From Inner Product with Zero Vector, we then have:

$\innerprod 0 y = 0$

so:

$\innerprod {n x} y = n \innerprod x y$

in this case.

Now we prove the statement for $n \ge 1$.

We proceed by induction.

For each $n \in \N$ let $\map P n$ be the proposition:

$\innerprod {n x} y = n \innerprod x y$

Basis for the Induction

We have:

$\innerprod {1 \times x} y = \innerprod x y$

So $\map P 1$ holds.

Induction Hypothesis

Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.

So this is our induction hypothesis:

$\ds \innerprod {n x} y = n \innerprod x y$

Induction Step

Now we need to show:

$\innerprod {\paren {n + 1} x} y = \paren {n + 1} \innerprod x y$

We have:

\(\ds \innerprod {\paren {n + 1} x} y\)

\(=\)

\(\ds \innerprod {n x + x} y\)

\(\ds \)

\(=\)

\(\ds \innerprod {n x} y + \innerprod x y\)

as show in Norm satisfying Parallelogram Law induced by Inner Product

\(\ds \)

\(=\)

\(\ds n \innerprod x y + \innerprod x y\)

from the induction assumption

\(\ds \)

\(=\)

\(\ds \paren {n + 1} \innerprod x y\)

So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$