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I'm preparing for my calculus exam. And I need two examples of infinite systems of functions with particular properties.

I call system $\Phi=\{\phi(x)\}$ ($\phi(x)$ is defined on $[a,b]$) complete if there is no function $f(x)\neq0,f(x)\in\Phi$ that satisfies $\int_a^bf(x)\phi(x)dx=0$, $\forall\phi(x)\in\Phi$.

I call system $\Phi=\{\phi_k(x)\}$ ($\phi_k(x)$ is defined on $[a,b]$ $\forall k\in \mathbb{N}$) closed if $\forall f(x)\in\Phi$, $\forall \varepsilon>0$ $\exists c_0,c_1,...,c_n\in\mathbb{R}$, that $||f-\sum_{k=0}^{n}c_k\phi_k(x)||<\varepsilon$.

So I need two examples for these definitions.

  • The examle of infinite system of functions (defined on $[-\pi,\pi]$), that is NOT closed in space $Q[-\pi,\pi]$, but it must be closed in subspace of $Q[-\pi,\pi]$, such that consisting of all even piecewise continuous functions.
  • The example of infinite orthogonal system of functions (defined on $[-\pi,\pi]$), that is not comlete in $Q[-\pi,\pi]$.

It would be great if you tell me about this systems or give a link.

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I think you made a mistake in the definition of closed. If $f \in \Phi$, then clearly $\|f-f\| = 0$ is a trivial solution. –  copper.hat Jan 10 '13 at 8:43
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For the first, take $\Phi = \{ x \mapsto \cos (nx) \}_{n=0}^\infty$. If $f$ is a piecewise continuous, even function, then it should be clear that $\int_{\pi}^\pi f(x) \sin (nx) dx = 0$ for all $n$, and hence the Fourier series for $f$ consists of $\cos$ terms only, hence $\Phi$ is closed in the set of piecewise continuous, even functions.

If $g$ is an even function, then we have $\max(|g(x)-x|,|g(x)+x|) \geq |x|$ for all $x$. Hence $\int_{-\pi}^\pi |g(x)-x|^2 dx = \int_0^\pi (|g(x)-x|^2+|g(x)+x|^2)dx \geq \int_0^\pi x^2 dx = \frac{\pi^3}{3}>0$. Consequently, if we let $\text{Id}(x) = x$, we see that $\|\text{Id} - \sum_k c_k \phi_k \| \geq \sqrt{\frac{\pi^3}{3}}$, for all $\phi_k \in \Phi$, for all choices of $c_k$, so $\Phi$ is not closed as per your definition.

For the second, take $\Phi$ as above. $\Phi$ is orthogonal, but the function $f(x) = \sin x$ is non zero, but $\int_{\pi}^\pi \sin x\, \phi(x) dx = 0$ for all $\phi \in \Phi$, hence $\Phi$ is not complete as per your definition.

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