Let $ V $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$.

It’s not hard to show that if $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ for some (real) inner product $\langle \cdot, \cdot \rangle$, then the parallelogram equality

$$ 2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u – v\rVert^2 $$

holds for all pairs $u, v \in V$.I’m having difficulty with the converse. Assuming the parallelogram identity, I’m able to convince myself that the inner product should be

$$ \langle u, v \rangle = \frac{\lVert u\rVert^2 + \lVert v\rVert^2 – \lVert u – v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 – \lVert u\rVert^2 – \lVert v\rVert^2}{2} = \frac{\lVert u + v\rVert^2 – \lVert u – v\rVert^2}{4} $$I cannot seem to get that $\langle \lambda u,v \rangle = \lambda \langle u,v \rangle$ for $\lambda \in \mathbb{R}$. How would one go about proving this?

**Answer**

Since this question is asked often enough, let me add a detailed solution. I’m not quite following Arturo’s outline, though. The main difference is that I’m not re-proving the Cauchy-Schwarz inequality (Step 4 in Arturo’s outline) but rather use the fact that multiplication by scalars and addition of vectors as well as the norm are continuous, which is a bit easier to prove.

So, assume that the norm $\|\cdot\|$ satisfies the parallelogram law

$$2 \Vert x \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y \Vert^2 + \Vert x – y \Vert^2$$

for all $x,y \in V$ and put

$$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 – \Vert x – y \Vert^2\right).$$ We’re dealing with *real* vector spaces and defer the treatment of the complex case to Step 4 below.

**Step 0.** $\langle x, y \rangle = \langle y, x\rangle$ and $\Vert x \Vert = \sqrt{\langle x, x\rangle}$.

Obvious.

**Step 1.** The function $(x,y) \mapsto \langle x,y \rangle$ is continuous with respect to $\Vert \cdot \Vert$.

Continuity with respect to the norm $\Vert \cdot\Vert$ follows from the fact that addition and negation are $\Vert \cdot \Vert$-continuous, that the norm itself is continuous and that sums and compositions of continuous functions are continuous.

*Remark.* This continuity property of the (putative) scalar product will only be used at the very end of step 3. Until then the solution consists of purely algebraic steps.

**Step 2.** We have $\langle x + y, z \rangle = \langle x, z \rangle + \langle y, z\rangle$.

By the parallelogram law we have

$$2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 = \Vert x + y + z \Vert^2 + \Vert x – y + z\Vert^2 .$$

This gives

$$\begin{align*}

\Vert x + y + z \Vert^2 & = 2\Vert x + z \Vert^2 + 2\Vert y \Vert^2 – \Vert x – y + z \Vert^2 \\

& = 2\Vert y + z \Vert^2 + 2\Vert x \Vert^2 – \Vert y – x + z \Vert^2

\end{align*}$$

where the second formula follows from the first by exchanging $x$ and $y$. Since $A = B$ and $A = C$ imply $A = \frac{1}{2} (B + C)$ we get

$$\Vert x + y + z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x + z \Vert^2 + \Vert y + z \Vert^2 – \frac{1}{2}\Vert x – y + z \Vert^2 – \frac{1}{2}\Vert y – x + z \Vert^2.$$

Replacing $z$ by $-z$ in the last equation gives

$$\Vert x + y – z \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2 + \Vert x – z \Vert^2 + \Vert y – z \Vert^2 – \frac{1}{2}\Vert x – y – z \Vert^2 – \frac{1}{2}\Vert y – x – z \Vert^2.$$

Applying $\Vert w \Vert = \Vert – w\Vert$ to the two negative terms in the last equation we get

$$\begin{align*}\langle x + y, z \rangle & = \frac{1}{4}\left(\Vert x + y + z \Vert^2 – \Vert x + y – z \Vert^2\right) \\

& = \frac{1}{4}\left(\Vert x + z \Vert^2 – \Vert x – z \Vert^2\right) +

\frac{1}{4}\left(\Vert y + z \Vert^2 – \Vert y – z \Vert^2\right) \\

& = \langle x, z \rangle + \langle y, z \rangle

\end{align*}$$

as desired.

**Step 3.** $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{R}$.

This clearly holds for $\lambda = -1$ and by step 2 and induction we have $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ for all $\lambda \in \mathbb{N}$, thus for all $\lambda \in \mathbb{Z}$. If $\lambda = \frac{p}{q}$ with $p,q \in \mathbb{Z}, q \neq 0$ we get with $x’ = \dfrac{x}{q}$ that

$$q \langle \lambda x, y \rangle = q\langle p x’, y \rangle = p \langle q x’, y \rangle = p\langle x,y \rangle,$$

so dividing this by $q$ gives

$$\langle \lambda x , y \rangle = \lambda \langle x, y \rangle \qquad\text{for all } \lambda \in \mathbb{Q}.$$

We have just seen that for fixed $x,y$ the continuous function $\displaystyle t \mapsto \frac{1}{t} \langle t x,y \rangle$ defined on $\mathbb{R} \smallsetminus \{0\}$ is equal to $\langle x,y \rangle$ for all $t \in \mathbb{Q} \smallsetminus \{0\}$, thus equality holds for all $t \in \mathbb{R} \smallsetminus \{0\}$. The case $\lambda = 0$ being trivial, we’re done.

**Step 4.** The complex case.

Define $\displaystyle \langle x, y \rangle =\frac{1}{4} \sum_{k =0}^{3} i^{k} \Vert x +i^k y\Vert^2$, observe that $\langle ix,y \rangle = i \langle x, y \rangle$ and $\langle x, y \rangle = \overline{\langle y, x \rangle}$ and apply the case of real scalars twice (to the real and imaginary parts of $\langle \cdot, \cdot \rangle$).

**Addendum.** In fact we can weaken requirements of Jordan von Neumann theorem to

$$

2\Vert x\Vert^2+2\Vert y\Vert^2\leq\Vert x+y\Vert^2+\Vert x-y\Vert^2

$$

Indeed after substitution $x\to\frac{1}{2}(x+y)$, $y\to\frac{1}{2}(x-y)$ and simplifications we get

$$

\Vert x+y\Vert^2+\Vert x-y\Vert^2\leq 2\Vert x\Vert^2+2\Vert y\Vert^2

$$

which together with previous inequality gives the equality.

**Attribution***Source : Link , Question Author : Hans Parshall , Answer Author : Norbert*