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[[File:Different Views on RKHS.png|thumb|right|Figure illustrates related but varying approaches to viewing RKHS]]
In [[functional analysis]]
An immediate consequence of this property is that convergence in norm implies [[uniform convergence]] on any subset of <math>X</math> on which <math>\|K_x\|</math> is bounded. However, the converse does not necessarily hold. Often the set <math>X</math> carries a topology, and <math>\|K_x\|</math> depends continuously on <math>x\in X</math>, in which case: convergence in norm implies uniform convergence on compact subsets of <math>X</math>.
It is not entirely straightforward to construct natural examples of a Hilbert space which are not an RKHS in a non-trivial fashion.<ref>Alpay, D., and T. M. Mills. "A family of Hilbert spaces which are not reproducing kernel Hilbert spaces." J. Anal. Appl. 1.2 (2003): 107–111.</ref> Some examples, however, have been found.<ref> Z. Pasternak-Winiarski, "On weights which admit reproducing kernel of Bergman type", ''International Journal of Mathematics and Mathematical Sciences'', vol. 15, Issue 1, 1992. </ref><ref> T. Ł. Żynda, "On weights which admit reproducing kernel of Szegő type", ''Journal of Contemporary Mathematical Analysis'' (Armenian Academy of Sciences), 55, 2020. </ref>
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:'''Theorem'''. Suppose ''K'' is a symmetric, [[positive definite kernel]] on a set ''X''. Then there is a unique Hilbert space of functions on ''X'' for which ''K'' is a reproducing kernel.
'''Proof'''. For all ''x'' in ''X'', define ''K<sub>x</sub>'' = ''K''(''x'', ⋅ ). Let ''H''<sub>0</sub> be the [[linear span]] of {''K<sub>x</sub>'' : ''x'' ∈ ''X''}. Define an inner product on ''H''<sub>0</sub> by
:<math> \left\langle \sum_{j=1}^n b_j K_{y_j}, \sum_{i=1}^m a_i K_{x_i} \right \rangle_{H_0} = \sum_{i=1}^m \sum_{j=1}^n {a_i} b_j K(y_j, x_i),</math>
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