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I am taking a course in Functional Analysis online, and unfortunately some important terms have not been well defined. In particular, isn't L2 space just Lp space with p=2 ? If so, why aren't continuous functions on closed intervals with the L2 norm Banach spaces (on finite dimensional spaces, by the Fischer-Riesz Theorem)?

PossumP
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    If you ignore the inner product, $L^2$ space is just $L^p$ space with $p = 2$. The inner product gives $L^2$ additional structure: it is not only a Banach space but also a Hilbert space. The subset of continuous functions in $L^p$ does not form a Banach space because it is not complete. –  Dec 12 '14 at 23:57
  • @Bungo you should post it as an answer. – TZakrevskiy Dec 13 '14 at 00:05
  • @Bungo Sorry, but I am confused about the definition of the terms. Doesn't Lp contain ALL functions that have a well defined Lebesgue integral over the interval in question? Also, I thought the inner product / norm was part of the definition of the space, and therefore cannot be ignored??? – PossumP Dec 13 '14 at 00:12
  • $L^p$ contains all functions for which $\int |f(x)|^p dx$ is well-defined and finite. If the domain is a closed bounded interval (or more generally, a compact subset of $\mathbb{R}$), then every continuous function is in $L^p$ for every $p$. But $L^p$ contains many other functions besides the continuous functions. –  Dec 13 '14 at 00:15
  • A function is in $L^2$ if and only if $\int |f(x)|^2 dx$ is well-defined and finite. This is true whether or not you pay any attention to the inner product. –  Dec 13 '14 at 00:17
  • Regarding the definitions: A Banach space is by definition a complete normed vector space. $L^2$ is a complete normed vector space which also has an inner product. So it is both a Banach space and a Hilbert space. For any $p$ (including $p=2$), the subset of $L^p$ consisting of the continuous functions is a normed vector space, but it is not complete, so it is not a Banach space. –  Dec 13 '14 at 00:21
  • Got it. Thanks. – PossumP Dec 13 '14 at 00:24

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If you ignore the inner product, $L^2$ space is just $L^p$ space with $p=2$. The inner product gives $L^2$ additional structure: it is not only a Banach space but also a Hilbert space.

The subset of continuous functions in $L^p$ (for $1 \leq p < \infty$) does not form a Banach space because it is not complete. However, it is a dense subset of $L^p$, meaning that given any $f \in L^p$, there is a sequence $f_n$ of continuous functions converging to $f$ in the $L^p$ norm. Putting it another way, an arbitrary $f \in L^p$ can be approximated arbitrarily closely (with respect to the $L^p$ norm) by continuous functions.

Regarding the definitions: A Banach space is by definition a complete normed vector space. $L^2$ is a complete normed vector space which also has an inner product. So it is both a Banach space and a Hilbert space. For any $p<\infty$ (including $p=2$), the subset of $L^p$ consisting of the continuous functions is a normed vector space, but it is not complete, so it is not a Banach space.