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.


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

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


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 $\langle y_{1},x\rangle = 0$.

  • 1
    $\begingroup$ I think we need more details here. What are the vectors $x_1,x_2$? $\endgroup$ – Mark Jul 11 '20 at 14:36
  • $\begingroup$ @Mark I have added the proof $\endgroup$ – SABOY Jul 11 '20 at 15:12

Note that $y_1, y_2, k$ are pairwise orthogonal, so by Pythagoras:


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 $\alpha_1y_1=k=0$.


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