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Let $\phi : \mathbb R^d \rightarrow \mathbb R^n$. Consider a kernel $K(x, y)$. Prove that for all $x, y \in \mathbb R^d, \| φ(x)−φ(y)\|_2 = \sqrt{K(x,x)−2K(x,y)+K(y,y)}.$

If I recall correctly $K(x, y)= ⟨\phi(x), \phi(y)⟩=\sum_{i=1}^{n}\phi_i(x)\phi_i(y)$, then $K(x,x)−2K(x,y)+K(y,y)=\sum_{i=1}^{n}(\phi_i^2(x)-2\phi_i(x)\phi_i(y)+\phi_i^2(y))=\sum_{i=1}^{n}(\phi_i(x)-\phi_i(y))^2=\\=| φ(x)−φ(y)\|_2^2 \Rightarrow \| φ(x)−φ(y)\|_2 = \sqrt{K(x,x)−2K(x,y)+K(y,y)}$.

Did I miss anything?

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