# Completeness of an inner product space

## Main Question or Discussion Point

I'm on a course which is currently introducing me to the concept of Hilbert spaces and the professor in charge was giving examples of such spaces. He ended by considering $$V$$, the space of polynomials with complex coefficients from $$\mathbb{R}$$ to $$\mathbb{C}$$. He then, for $$f,g\in V$$, defined

$$(f,g)=\int_{0}^{\infty}f(x)\bar{g(x)}e^{-x}dx$$​

and claimed - without proof - that $$V$$ equipped with $$(\cdot ,\cdot )$$ is an inner product space, but that $$V$$ is not complete. Could anyone come up with a clever way of showing that this is true?

mathman
There is a classical theorem which applies here (from Wikipedia).
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In mathematical analysis, the Weierstrass approximation theorem states that every continuous function defined on an interval [a,b] can be uniformly approximated as closely as desired by a polynomial function. Because polynomials are among the simplest functions, and because computers can directly evaluate polynomials, this theorem has both practical and theoretical relevance, especially in polynomial interpolation. The original version of this result was established by Karl Weierstrass in 1885.
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In other words, (not polynomial) continuous functions can be the limit of polynomials. Therefore the space of only polynomials is not complete.

You would need to do a little work to extend this to the space of functions you are concerned with, but the general idea would be to approximate non - polynomials with polynomial sequences using the given inner product norm, rather than uniform approximation (sup norm).

Thank you for your help, I'll get on it later today.

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Edit: Rubbish.

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mathman
Let f(x)=e-x and fn(x) be the polynomial defined by the first n terms. This will show it is not complete, since fn(x) converges to f(x), which is not a polynomial. Moreover, it is easy to show that any polynomial will have a finite norm.

HallsofIvy
of a Taylor's series expansion of $e^{-x}$.