The Parallelogram Law | Real Analysis

An identity hides between the two diagonals and the two sides of a parallelogram. This scene slides the exponent

p

and checks when that identity survives.

Every norm that comes from an inner product satisfies the identity. Computing with

e_1

and

e_2

makes the left side

2 \cdot 2^{2/p}

and the right side

4

. The two agree only at

p = 2

. So an

\ell^p

norm with

p

different from 2 cannot come from any inner product.

For example, a 📐 parallelogram drawn with a ruler stays exactly the same, but the measuring ruler is swapped. If swapping the ruler alone breaks the diagonal identity, that ruler never came from an inner-product blueprint.

Examples & Exercises

Prove completely that in a real inner product space, every $x, y$ satisfies the parallelogram law $\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2$ .

\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2

p = 2

: the Euclidean norm comes from an inner product, and the left side

= 4 =

the right side — the case where the parallelogram law holds.

Show proof

Expand the norm of the sum through the inner product. In a real inner product space $\|x+y\|^2 = \langle x+y, x+y \rangle$ , and bilinearity with symmetry gives $\langle x+y, x+y \rangle = \langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle = \|x\|^2 + 2\langle x, y \rangle + \|y\|^2$ . That is, $\|x+y\|^2 = \|x\|^2 + 2\langle x, y \rangle + \|y\|^2$ .
Expand the norm of the difference the same way. $\|x-y\|^2 = \langle x-y, x-y \rangle = \langle x, x \rangle - \langle x, y \rangle - \langle y, x \rangle + \langle y, y \rangle = \|x\|^2 - 2\langle x, y \rangle + \|y\|^2$ . That is, $\|x-y\|^2 = \|x\|^2 - 2\langle x, y \rangle + \|y\|^2$ .
Add the two identities. The cross terms cancel: $\|x+y\|^2 + \|x-y\|^2 = \left(\|x\|^2 + 2\langle x, y \rangle + \|y\|^2\right) + \left(\|x\|^2 - 2\langle x, y \rangle + \|y\|^2\right) = 2\|x\|^2 + 2\|y\|^2$ . This is the parallelogram law, and the proof is complete.
Remark: this is the real case. The law itself also holds in complex inner product spaces by the same expansion, but the polarization identity in problem 3 below changes to a form using $\mathrm{Re}$ there — not needed here.

For $e_1 = (1,0)$ and $e_2 = (0,1)$ in $\ell^p(\mathbb{R}^2)$ , compute both sides of the parallelogram law and prove completely that equality holds only when $p = 2$ . Conclude that for $p \ne 2$ the $\ell^p$ norm is not induced by any inner product.

\|e_1+e_2\|_p^2 + \|e_1-e_2\|_p^2 = 2^{1 + 2/p}, \qquad 2\|e_1\|_p^2 + 2\|e_2\|_p^2 = 4 = 2^2

p = 3

: the left side

= 2^{5/3} \approx 3.1748

, the right side

= 4

— the parallelogram law visibly fails.

Show proof

Compute each norm first. Since $e_1 + e_2 = (1,1)$ , $\|e_1+e_2\|_p = \left(|1|^p + |1|^p\right)^{1/p} = 2^{1/p}$ , and squaring gives $\|e_1+e_2\|_p^2 = \left(2^{1/p}\right)^2 = 2^{2/p}$ . Likewise $e_1 - e_2 = (1,-1)$ , so $\|e_1-e_2\|_p = \left(|1|^p + |-1|^p\right)^{1/p} = 2^{1/p}$ and $\|e_1-e_2\|_p^2 = 2^{2/p}$ .
Collect both sides. The left side is $\|e_1+e_2\|_p^2 + \|e_1-e_2\|_p^2 = 2^{2/p} + 2^{2/p} = 2 \cdot 2^{2/p} = 2^{1 + 2/p}$ . Meanwhile $\|e_1\|_p = \|e_2\|_p = 1$ , so the right side is $2\|e_1\|_p^2 + 2\|e_2\|_p^2 = 2 \cdot 1 + 2 \cdot 1 = 4 = 2^2$ . Hence equality is equivalent to $2^{1 + 2/p} = 2^2$ .
$(\Rightarrow)$ Suppose equality $2^{1 + 2/p} = 2^2$ holds. The map $t \mapsto 2^t$ is injective, so the exponents must agree: $1 + 2/p = 2$ . Hence $2/p = 1$ , i.e. $p = 2$ .
$(\Leftarrow)$ Conversely, suppose $p = 2$ . Then $1 + 2/p = 1 + 1 = 2$ , so the left side is $2^{1 + 2/p} = 2^2 = 4$ , directly equal to the right side $4$ . Hence equality holds.
Conclusion: for $p \ne 2$ the $\ell^p$ norm violates the parallelogram law. Since problem 1 shows that every inner-product norm satisfies the law, by the Jordan-von Neumann theorem the $\ell^p$ norm for $p \ne 2$ is not induced by any inner product.

Verify completely that in a real inner product space the polarization identity $\langle x, y \rangle = \frac{\|x+y\|^2 - \|x-y\|^2}{4}$ holds.

\langle x, y \rangle = \frac{\|x+y\|^2 - \|x-y\|^2}{4}

p = 2

: the Euclidean case, where the parallelogram law holds — the polarization identity reads the inner product back off the norm.

Show proof

Reuse the expansions from problem 1: $\|x+y\|^2 = \|x\|^2 + 2\langle x, y \rangle + \|y\|^2$ and $\|x-y\|^2 = \|x\|^2 - 2\langle x, y \rangle + \|y\|^2$ .
Subtract the two identities. The $\|x\|^2$ and $\|y\|^2$ terms disappear and only the cross terms survive: $\|x+y\|^2 - \|x-y\|^2 = \left(\|x\|^2 + 2\langle x, y \rangle + \|y\|^2\right) - \left(\|x\|^2 - 2\langle x, y \rangle + \|y\|^2\right) = 4\langle x, y \rangle$ .
Divide both sides by $4$ : $\frac{\|x+y\|^2 - \|x-y\|^2}{4} = \langle x, y \rangle$ — the polarization identity is verified. The inner product is completely determined by the norm alone.
By the Jordan-von Neumann theorem, whenever a norm satisfies the parallelogram law, this very formula defines an inner product for it (statement only).