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Let $0\neq f \in L^2(\mathbb R)$ (complex Hilbert space).

(1) Can we say that $\left\{ f(x), e^{2\pi i x }f(x) \right\}$ is Linearly independent in $L^2(\mathbb R)$?

(2) Is there any known method(s) one should try to show finite set (consists of finite vectors) is Linearly independent? If so, I'm curios to know existing methods to show the finite set is linearly independent in $L^2(\mathbb R)$?

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put on hold as unclear what you're asking by Did, Cesareo, Riccardo.Alestra, José Carlos Santos, A. Pongrácz Jan 17 at 10:53

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    $\begingroup$ Can you write one as a scalar multiple of the other with pointwise a.e. equality? $\endgroup$ – David Bowman Jan 12 at 20:04
  • $\begingroup$ Do you mean linearly independent over $\mathbb{C}$ (then the answer is yes) or do you mean over $L^2(\mathbb{R})$ (then the answer is no)? $\endgroup$ – Severin Schraven Jan 13 at 0:26
  • $\begingroup$ @SeverinSchraven: Thanks. I mean, LI over $\mathbb C$. Can explain bit more how to conclude this? Thanks. $\endgroup$ – Math Learner Jan 14 at 16:23
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Assume we have $\lambda, \mu\in\mathbb C$ such that $$\lambda f(x) + \mu e^{i2\pi x} f(x) =0$$ Let $$A=\{x\in\mathbb R\ :\ f(x) \neq 0\}$$ We know that $A$ has positive measure, otherwise we have $f=0$ a.e. For all $x\in A$ we have $$\lambda + \mu e^{i2\pi x} = 0$$ We know for this to be true we should have $\mu=\lambda e^{i\theta} $ for some $\theta$ (why?). Then $$\lambda (1+e^{i(2\pi x+\theta)}) =0$$ Either $\lambda = 0$ or $(1+e^{i(2\pi x+\theta)})=0$. The second equality can only hold for countable many $x$ (why?), but we knew the measure of $A$ was positive so $A$ is not countable. Hence we have $\lambda=0$ and therefore also $\mu=0$ proving the independence of the given set.

For such questions there are multiple ways to approach. I know that using the definition as above helps. You can also show that in an inner product space as $L^2$ you might get an orthonormal set which is also independent.

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