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How to find an orthonormal set in the space of compactly supported smooth functions on $\mathbb{R}$? Moreover, for the operator $\frac{d^2}{dt^2}$, what are the eigenfunctions in the space of compactly supported smooth functions on $\mathbb{R}$?

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    $\begingroup$ I realize you wrote "an orthonormal set" rather than "basis", but finding just an orthonormal set is too easy: take one smooth compactly supported function, and normalize it. Voilà, an orthonormal set with 1 element. Throw in its translates with disjoint supports to get an infinite orthonormal set. $\endgroup$ – user357151 May 10 '17 at 3:32
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Orthonormal basis in the space of compactly supported smooth functions

One has to define an inner product first; I assume you mean the $L^2$ inner product, $\langle f,g\rangle = \int_{\mathbb{R}} fg$. Since normalizing the basis vectors is easy, let's focus on orthogonal bases.

The Daubechies wavelet provides an orthogonal basis for $L^2(\mathbb{R})$ which consists of compactly supported functions of class $C^{k}$, where $k$ can be made arbitrarily large (at the cost of increasing the complexity of the wavelet). See sections 1.3 and 1.4 of these lecture notes.

There does not exist a $C^\infty$-smooth compactly supported orthogonal wavelet. Thus, the typical options for an explicit orthogonal basis in $L^2(\mathbb{R})$ consist of the following.

  • Give up compact support and use Hermite functions or Meyer wavelet.
  • Downgrade from infinite smoothness to $C^k$ smoothness and use the Daubechies wavelet or its relatives.
  • Give up all smoothness (and even continuity) and use the much simpler Haar wavelet or the functions $f_{m,n}(t) = \exp(2\pi i n t)\chi_{[m,m+1]}$.

In principle, an orthonormal basis consisting of $C^\infty$ smooth compactly supported functions can be constructed by applying the Gram-Schmidt process to some complete set of such functions (like monomials multiplied by the shifts of a smooth bump function). But the computation would be intractable, and the resulting basis not even remotely explicit.

Smooth compactly supported eigenfunctions of $d^2/dx^2$

There are none. On the Fourier transform side, applying $d^2/dx^2$ amounts to multiplying by $-\xi^2$. So the Fourier transform of an eigenfunction has to be supported on the set $\{\pm \sqrt{-\lambda}\}$, which means the eigenfunction is of the form $A\cos \sqrt{-\lambda} x+B\sin \sqrt{-\lambda} x$. Not compactly supported, obviously.

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  • $\begingroup$ thanks a lot for the explanation, yes I was looking for a basis!! $\endgroup$ – Sanand May 11 '17 at 6:56

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