# Norms Induced by Inner Products and the Parallelogram Law

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?

• I added "Parallelogram Law" to the title in order to make this question a bit easier to find for people wanting to ask about this. I hope you don't mind. – t.b. Jul 2 '11 at 1:24
• You should also check that the induced objects do correspond to the original objects and not something entirely new (that is important so I added an answer refering to this though it is not e plicitely the question you were asking). – C-Star-W-Star Aug 24 '14 at 17:35
• This is the famous Von Neumman-Fréchet theorem. – Joshua Benabou Oct 25 '17 at 10:00

It's not immediate or trivial, so I wouldn't feel too bad for having trouble. This is an exercise in Friedberg, Insel, and Spence's Linear Algebra, 4th Edition, which has an extensive 8 part "Hint." Here's an edited sequence of hints, following theirs:

1. First, prove that the result holds for $\lambda = 2$, that is, $\langle 2u,v\rangle = 2\langle u,v\rangle$.

2. Then, prove that the inner product is additive in the first component: $\langle x+u,v\rangle = \langle x,v\rangle + \langle u,v\rangle$.

3. Then, prove the result holds for $\lambda$ any positive integer. Then for the reciprocal $\frac{1}{m}$ of any positive integer. Then for any rational number.

4. Then prove that $|\langle u,v\rangle|\leq ||u||\,||v||$

5. Then prove that for every $\lambda\in\mathbb{R}$, every $r\in\mathbb{Q}$, you have $$|\lambda\langle u,v\rangle - \langle \lambda u,v\rangle | = |(\lambda-r)\langle u,v\rangle - \langle(\lambda-r)u,v\rangle|\leq 2|\lambda-r|\,||u||\,||v||.$$

6. Finally, use that to prove homogeneity: for every $\lambda\in\mathbb{R}$, $\langle\lambda u,v\rangle = \lambda\langle u,v\rangle$.

• I added a rather detailed solution to this problem, as this gets asked often enough. Since your explanatory and expository skills surpass mine by light years, I'd appreciate it if you could go over the solution and clarify and simplify it wherever you feel the need. – t.b. Jun 7 '11 at 10:25
• Hi, do you know by any chance if the proof of this fact over Complex numbers is similar? – Vadim Sep 15 '15 at 20:23

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.

• @all: Feel free to edit this answer in case you find anything unclear or should be improved. I'm not making the answer community wiki to make sure that no errors get edited in (relying on the peer review process). – t.b. Jun 7 '11 at 10:19
• I should have said that I copied this argument more or less faithfully from approximately 10 year old notes of mine. I'm pretty sure that these notes were worked out following a book but I couldn't reconstruct which one. – t.b. Jun 7 '11 at 15:06
• @Theo Excellent writeup! – ItsNotObvious Jun 7 '11 at 22:35
• @3Sphere: Thank you, that's very kind of you! – t.b. Jun 7 '11 at 22:46
• @Theo Seems pretty clear to me. With respect to continuity, it might not hurt to add the the norm itself is Lipschitz and therefore continous – ItsNotObvious Jun 8 '11 at 1:49

One should also convince oneself that: $$\langle\cdot,\cdot\rangle\to\|\cdot\|\to\langle\cdot,\cdot\rangle$$ $$\|\cdot\|\to\langle\cdot,\cdot\rangle\to\|\cdot\|$$ (Otherwise really bad things could happen...)

Luckily, this can be checked rather easily: $$\|x\|'=\sqrt{\frac{1}{4}\left(\|x+x\|^2-\|x-x\|^2\right)}=\|x\|$$ $$\langle x,y\rangle'=\frac{1}{4}\left(\sqrt{\langle x+y,x+y\rangle}^2-\sqrt{\langle x-y,x-y\rangle}^2\right)=\langle x,y\rangle$$

• I don't understand your point -- what kind of really bad things do you have in mind? – Oskar Henriksson Nov 21 '17 at 11:22
• An "applied" example: Consider you have Hilbert space with scalar product $\langle\cdot,\cdot\rangle$. You define the norm $\|\cdot\|:=\sqrt{\langle\cdot,\cdot\rangle}$. Now somebody gives you an operator $J\in B(H)$ from some application and that person tells you he has checked that $\|Jx\|=\|x\|$ for all $x\in H$. He asks you to find out wether it follows already that $\langle Jx,Jy\rangle=\langle x,y\rangle$ for all $x,y\in H$... – C-Star-W-Star Nov 21 '17 at 14:42
• ... So you ambitiously start your work as mathematician and apply polarization to get $\langle Jx,Jy\rangle'=\langle x,y\rangle'$ for all $x,y\in H$. But wait, whoops, how can you be sure that this was the original scalar product, i.e. why should it hold true that $\langle\cdot,\cdot\rangle'=\langle\cdot,\cdot\rangle$? – C-Star-W-Star Nov 21 '17 at 14:43
• As a note: This is a mistake one can trap into quite easily and actually quite often. For example something more advanced would be, start with a locally compact but non-Hausdorff space $X$, consider the continuous complex functions vanishing at infinity, $C_0(X)$. Findout some C*-algebraic properties and deduce the analogue for the topological space via Gelfand transformation. Whoops, trap, the Gelfand representation returns a topological space that is not isomorphic to the original space. – C-Star-W-Star Nov 21 '17 at 14:53

(sorry, I'm not familiar with LaTex, hope you have no difficutly reading my answer).

I have another method to prove this result for $\ x,y \in \mathbb{R^n}$, $\lambda \in \mathbb{R}$.

step.1. prove that $\langle \lambda x,\lambda y \rangle = \lambda^2 \langle x, y \rangle$, use polarisation identity to expand inner product, it's easy to prove it.

step.2. prove that $\langle \lambda x, y \rangle = \langle x, \lambda y \rangle$, i.e.prove $\langle \lambda x, y \rangle - \langle x, \lambda y \rangle=0$, the proof is similar to proving $\langle x+y, z \rangle = \langle x, z \rangle+\langle y, z \rangle$, you may use $\langle x+y, z \rangle = \langle x, z \rangle+\langle y, z \rangle$ and $\langle -x, y \rangle = -\langle x, y \rangle=\langle x,-y \rangle$ in your proof and the latter is easy to prove.

step.3. use the result of step.2., we have $\langle \lambda x,\lambda y \rangle = \langle \lambda^2 x, y \rangle$, compare to the result in step.1., we get that $\langle \lambda^2 x , y \rangle = \lambda^2 \langle x, y \rangle$, for $\lambda^2 >0$.

step.4 when $\lambda <0$, $\langle \lambda x, y \rangle = \langle-(-\lambda x), y \rangle=-\langle(-\lambda x), y \rangle$, use the result in step.3., we get $\langle(-\lambda x), y \rangle=-\lambda\langle x, y \rangle$, which means the result holds for $\lambda <0$, and it's easy to prove when $\lambda =0$.

It's a bit more complex to prove for $\ x,y \in \mathbb{C^n}, \lambda \in \mathbb{C}$, while the method is general, you just have to seperate it into two parts, the real and the imaginary.

• Note that the inappropriate references to $\Bbb Q$ have been changed (not by me) to $\Bbb C$, which makes much more sense here. If you really meant $\Bbb Q$ just change it back. – Rory Daulton Feb 23 '16 at 11:38
• I don't understand the I have another method'' part. Is not step 2 of this answer the same as the step 2 + step 3 of t.b.? – Hashimoto Feb 18 '19 at 7:14
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