# Under what circumstances does $\lvert \lambda\rvert \lvert \lvert x \rvert \rvert = \lvert \lvert y \rvert \rvert$ imply $\lambda x = y$

In a proof I saw, we made use of the fact that

for some $$y = \lambda_{1}x_{1} + \lambda_{2}x_{2}$$ , if we have $$\lvert \lambda_{1}\rvert \lvert \lvert x_{1} \rvert \rvert = \lvert \lvert y \rvert \rvert$$, then we can conclude that $$y = \lambda_{1}x_{1}$$.

This can certainly not be true in a general case, right? What assumptions are needed for it to be true. In the proof I mentioned, the space we were investigating is a Hilbert space.

Edit:

Let $$\mathcal{H}$$ be a Hilbert space, and $$F_{1},F_{2}$$ two bounded linear functionals such that $$F_{1}\neq 0$$ and $$F_{2}\neq 0$$. Suppose that

$$\forall x \in \mathcal{H}: \lvert F_{1}(x)\rvert=\lvert \lvert F_{1}\rvert \rvert \cdot \lvert\lvert x \rvert \rvert \implies F_{2}(x)=0$$

Now show that

$$\forall x \in \mathcal{H}: \lvert F_{2}(x)\rvert=\lvert \lvert F_{2}\rvert \rvert \cdot \lvert\lvert x \rvert \rvert \implies F_{1}(x)=0$$

Proof:

Identify, $$y_{1},y_{2}$$ with $$F_{1}(\cdot)=\langle y_{1}, \cdot\rangle$$ and $$F_{2}(\cdot)=\langle y_{2}, \cdot\rangle$$ by Riesz representation, then we can clearly see that:

$$\lvert F_{1}(y_{1})\rvert=\langle y_{1},y_{1}\rangle = \lvert \lvert y_{1}\rvert \rvert^{2}\implies \langle y_{2},y_{1}\rangle = 0$$

Now consider the closed subspace $$K:=\operatorname{span}\{y_{1},y_{2}\}$$. Then, by the orthogonal projection theorem, every $$x \in \mathcal{H}$$ can be written as $$x = \alpha_{1}y_{1}+\alpha_{2}y_{2}+k$$ where $$k \in K^{\perp}$$

And hence we assume that for some $$x \in \mathcal{H}$$ that $$\lvert \langle y_{2}, x\rangle \rvert= \lvert \lvert y_{2}\rvert \rvert \cdot \lvert \lvert x \rvert \rvert$$. Note that

$$\langle y_{2},x\rangle = \lvert \alpha_{2} \rvert \cdot \lvert \lvert y_{2}\rvert \rvert^{2}\implies \lvert \alpha_{2} \rvert \cdot \lvert \lvert y_{2}\rvert \rvert^{2}=\lvert \lvert x \rvert \rvert \cdot \lvert \lvert y_{2}\rvert \rvert\implies \lvert \lvert x \rvert \rvert =\lvert \alpha \rvert \cdot \lvert \lvert y_{2}\rvert \rvert$$

And then the implication which I do not understand is stated:

$$x = \alpha_{2}y_{2}$$, hence implying that $$\langle y_{1},x\rangle = 0$$.

• I think we need more details here. What are the vectors $x_1,x_2$? – Mark Jul 11 '20 at 14:36
• @Mark I have added the proof – SABOY Jul 11 '20 at 15:12

Note that $$y_1, y_2, k$$ are pairwise orthogonal, so by Pythagoras:
$$||x||^2=||\alpha_1y_1||^2+||\alpha_2y_2||^2+||k||^2$$
But during the proof we also got that $$||x||^2=||\alpha_2y_2||^2$$. Hence we must have $$||\alpha_1y_1||^2+||k||^2=0$$, which implies $$\alpha_1y_1=k=0$$.