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Am studying the basic concepts of RKHS and the representer theorem: In $f(x_i)=$, what does $ f$ on the r.h.s denote? What is its structure-is it a vector? I was thinking that $k(x_i,\mathbb{.})$ would be a vector and $f$ should be one. Some formal-yet simplified explanation would help me get a head-start in my reading!

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    Also, I was assuming that $k(x_i,\mathbb{.})$ is a vector formed by the placeholder taking in each entry in a vector $x$ except for $x_i$. Is my understanding right about this?2012-10-20

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I might be misunderstanding the question here, but it seems that you're a bit confused about what $$ means. Yes, $f$ is a vector, but it's not necessarily a column vector like you might be used to from linear algebra. It is a vector in some Hilbert space, and the inner product there could likely be something like an integral (e.g., $ = \int_{-\infty}^{\infty} f(x) k(x_i,x) dx$).

Maybe an example might help. Consider the space $H^1(\mathbb{R})$, which is the space of functions $\{f \in L^2: f' \in L^2\}$ which has an inner product$_{H^1} = \int_{-\infty}^{\infty} f(x) \bar{g}(x)dx + \int_{-\infty}^{\infty} f'(x) \bar{g'}(x) dx$

Note that the vectors in this space are functions, and not just a simple column vector. This space has a reproducing kernel, namely $k(x,y) = \frac{1}{2} e^{|x-y|}$. The reproducing property tells us that we can 'sample' $f$ at any point we want by taking the inner product of $f$ with $k(x,\cdot)$. That is, if you give me a function $f$ and you want to know its value at some point, say $2$, then the reproducing property tells us that:

$f(2) = _{H^1} = \int_{-\infty}^{\infty} f(y) \frac{1}{2} e^{-|2-y|} + f'(y) \frac{d}{dy} \frac{1}{2} e^{-|2-y|} dy$

For a reference, check out Scattered Data Approximation by Holger Wendland, chapter 10.