# If $\langle x,y\rangle = \langle x,z\rangle$ for all $x \in H$ then $y=z$

My professor made the following claim in the body of a proof and verbally explained why it was justified. I thought I understood his explanation at the time but on reviewing my notes it's not as clear. Here is the claim:

$$\text{If } \ \langle x,y\rangle = \langle x,z\rangle \ \text{ for all } \ x \in H \ \text{ then } \ y=z$$

Here, $H$ is a Hilbert Space and $f(x)=\langle x,y\rangle$ for all $x \in H$ is a bounded linear functional.

The claim seems to be common sense, but I'd like mathematical justification. We were talking about the Riesz representation theorem for most of the class so I'm guessing it factors in here.

• Write it as $\langle x, y-z\rangle = 0$ for all $x$ and choose $x = y-z$. – Daniel Fischer Mar 8 '14 at 23:23
• Please consider using \langle and \rangle to get $\langle x,y\rangle$ instead of < and > to get $<x,y>$. – Fly by Night Mar 8 '14 at 23:28
• Thanks, I can see how choosing x = y - z gives us y = z. But what about in the case where x $\neq$ y - z? Or is x = y - z necessarily (since the Riesz representation theorem implies uniqueness). – Zachary Mar 8 '14 at 23:42

Expanding on Daniel's comment: suppose that $\langle x,y\rangle = \langle x,z\rangle$. Then we must have $$\langle x,y-z \rangle = \langle x,y \rangle - \langle x,z \rangle = 0$$ For all $x$. It certainly follows that, in the instance that $x = y-z$, we have $\langle x,y-z \rangle = 0$. Thus, we have $$\|y-z\| = \sqrt{\langle y-z,y-z \rangle} = 0$$ Because $\|\cdot \|$ is a norm, we have $\|x\| \implies x = 0$. Thus, we conclude that $y-z = 0$, which is to say that $y = z$.

No Riesz representation required.

• This is clear, thank you. I'm just wondering why we cannot have x $\neq$ y - z. Since equality holds for all x. – Zachary Mar 9 '14 at 0:01
• @user109844 It holds for all $x$ indeed, so in particular for $x=y-z$. We're using the fact that $\langle x, y-z\rangle=0$ holds for all $x$, not proving it. We choose $x$ the way we do because it suits our needs, no need to use all possible values. – Marcin Łoś Mar 9 '14 at 0:28
• I managed to confuse myself somehow. I understand what you're saying. Thanks to you and everyone else. – Zachary Mar 9 '14 at 0:36

Let $\delta(z) = \langle x-y,z\rangle$ and $L:\to H'$ be the Riezs bijection: $$L(a)(z) = \langle a,z\rangle$$ As $L(x-y) = \delta = 0_{H'} = L(0_H)$, $x=y$.

• Please consider using \langle and \rangle to get $\langle x,y\rangle$ instead of the inequality symbols < and > which give $<x,y>$. – Fly by Night Mar 8 '14 at 23:30
• ok. thank you for the trick. – mookid Mar 8 '14 at 23:32

Also you can think of it this way.The only vector in the Hilbert space $\mathcal{H}$ that is perpendicular to every single vector of $\mathcal{H}$ is zero.