I am always confused about what norm or inner product should I use given a specific space.
When talking about continuously differentiable functions, we mean $C^1$ not $L_{2}$ functions right?
Then can I use the norm $<f,g>=\int f(x)\overline{g(x)}dx$ which I think refers to the $L_{2}$ only (is it?) here? And if I can, is it because all norms (so inner product) are equivalent?
I have this question because I was trying to use the theorem:
An orthonormal sequence is complete iff $\sum|<x,e_{k}>|^2=||x||^2$.
And if the orthonormal sequence is complete, we then have $<x,y>=\sum <x,e_{k}>\overline{<y,e_{k}>}$.
And I already known that $e_{k}$ is complete in $H=L_{2}$, but the reason I'm hesitating is that the functions are 'continuously differentiable fucntions', which I think should be $C^1$ but not $L_{2}$.
Thanks for help!
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1$\begingroup$ Your inner product seems to work only for continuous(ly differenetiable) functions with compact support $\endgroup$– Hagen von EitzenFeb 26, 2017 at 21:35
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$\begingroup$ @HagenvonEitzen the continuously differentiable functions are on closed interval [-pi,pi] and are periodic, but I think this still cannot guarantee compact support right? So which means I cannot use this inner product here? But then what is the common inner product we are referring to on continuously differentiable functions on closed interval? $\endgroup$– LazerFeb 26, 2017 at 21:40
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1$\begingroup$ @J.Y It depends on context. The two most common that come to mind are the $L^2$ inner product and the Sobolev $H^1$ (aka $W^{1,2}$) inner product. There is no inner product that induces the $C^1$ norm that I reference in my answer. $\endgroup$– IanFeb 26, 2017 at 21:42
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1$\begingroup$ @J.Y: Are you studying Fourier series? If this is the context, then the inner product which you work with is probably the $L^2$ inner product over $[-\pi,\pi]$ (that is, $\left< f, g \right> := \int_{-\pi}^{\pi} f(x) \overline{g(x)} \, dx$). Just take note that $C^1_{\text{periodic}}([-\pi,\pi])$ together with $\left< \cdot, \cdot \right>$ is not a Hilbert space because it is not complete. $\endgroup$– levapFeb 26, 2017 at 21:47
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$\begingroup$ @levap Thanks for your help! Yes, when studying orthonormal sequence in functional analysis. And when given a space, it seems that we just pick certain inner products. But I remember that when talking about inner product and norm, we usually use the sup norm for C space, and it seems widely true that norm=sqrt(inner product) in our problems, but it makes more sense to use L2 inner product on the problem I'm working on, so I got confused LOL. $\endgroup$– LazerFeb 26, 2017 at 21:55
2 Answers
Not all norms are equivalent in infinite dimensional spaces. This means that, for example, $C^1(D)$ with the "$C^1(D)$ norm", i.e. $\| f \|_{C^1(D)} = \sup_{x \in D} |f(x)| + \sup_{x \in D} |f'(x)|$, is different from $C^1(D)$ with the $L^2$ norm. For one thing, the former is complete (as a metric space) while the latter is not. This means that in infinite dimensions, even if all you care about is topology, you must say something about the norm that you are using in order to properly make sense of your statements.
One important fact is that $C^1$ embeds into $L^2$ if the domain is bounded. Thus if you have a complete orthonormal sequence on a bounded domain, then it is also "complete for $C^1$", i.e. you can expand any $C^1$ function in terms of the sequence.
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$\begingroup$ Thanks for your answer! So whether an orthonormal sequence is complete or not has nothing to do with the inner product or norm I choose right? Only the space matters? $\endgroup$– LazerFeb 26, 2017 at 21:58
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1$\begingroup$ @J.Y No, it does matter, because the convergence of the expansion is in the norm induced by the inner product. $\endgroup$– IanFeb 26, 2017 at 22:13
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$\begingroup$ Maybe I misunderstood, what do you mean by 'C1 embeds into L2', is it 'C1 is a subset of L2'? but I guess you don't mean this.. $\endgroup$– LazerFeb 26, 2017 at 22:21
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1$\begingroup$ @J.Y "Embeds into" usually means "is a subset of and the inclusion map is nice". In this case the inclusion map is continuous. $\endgroup$– IanFeb 26, 2017 at 22:29
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The inner product/norm on a specific vector space should be defined explicitly or, as is more often the case, understood implicitly from the context. For example, consider the space
$$ C^1([a,b]) := \{ f \colon [a,b] \rightarrow \mathbb{C} \, | \, f \text{ is continuously differentiable} \}. $$
Without any context, this is just a vector space. Two possible structures you can put on it are:
- The $L^2$-inner product defined by $$ \left< f, g \right>_{L^2} := \int_a^b f(x) \overline{g(x)} \, dx. $$ Using this inner product, we have a notion of norm $\| f \|_{L^2} := \sqrt{\left< f, f \right>_{L^2}}$. Now, $C^1([a,b])$ together with $\left< \cdot, \cdot \right>_{L^2}$ is an inner produt space but it is not complete so it is not a Hilbert space.
- The $C^1$ norm is defined by $$ \| f \|_{C^1} := \sup_{x \in [a,b]} |f(x)| + \sup_{x \in [a,b]} |f'(x)|. $$ The space $C^1([a,b])$ together with $\| \cdot \|_{C^1}$ is a complete normed space (a Banach space) but it is not a Hilbert space (we haven't defined an inner product and in fact, it is not possible to define an inner product whose associated norm is $\| f \|_{C^1}$).
Which structure you care about and use depends on the context. In many (but not all) cases one wants to work with a complete vector spaces so it often makes more sense to consider $(C^1([a,b]), \| \cdot \|_{C^1})$ than ($C^1([a,b]), \left< \cdot, \cdot \right>_{L^2})$.
