I am reading S. Shimorin's paper titled Complete Nevanlinna-Pick property of Dirichlet-type spaces. My question concerns Lemma 2.3. which is as follows:
Assume $\mathscr{H}$ is a Hilbert space of functions defined on a set $\mathscr{X}$ with the norm $\lVert \cdot \rVert$ and the reproducing kernel $k$. Assume also that the Hilbert norms $\lVert \cdot \rVert_{n}, n \geqslant 0$ in $\mathscr{H}$ are uniformly equivalent to the norm $\lVert \cdot \rVert$ and satisfy \begin{equation*} \lvert \lVert x \rVert _n ^2 - \lVert x \rVert_0 ^2 \rvert \le \varepsilon_n \lVert x \rVert ^2, \qquad x \in \mathscr{H} \end{equation*} for some sequence $\varepsilon_n \to 0$. Then we have \begin{equation*} \lim_{n \to \infty} k^{n} (z, \lambda) =k^{0} (z,\lambda) \end{equation*} pointwise in $\mathscr{X} \times \mathscr{X}$, where $k^{n}, n \geqslant 0$ is the reproducing kernel for $\mathscr{H}$ with respect to the norm $\lVert \cdot \rVert_n$.
Although Shimorin does not define what a sequence of norms being uniformly equivalent to some norm mean but here's what I presume: A sequence of norms $(\|\cdot\|_n)_{n=1}^\infty$ on a vector space $X$ is said to be uniformly equivalent to a norm $\|\cdot\|$ if there exist positive constants $C_1$ and $C_2$ such that for all $x \in X$ and for all $n$:
$$C_1 \|x\| \leq \|x\|_n \leq C_2 \|x\|.$$
My question is: Is the uniform equivalence of the $\lVert \cdot \rVert_{n}, n \geqslant 0$ in $\mathscr{H}$ to the norm $\lVert \cdot \rVert$ in Lemma required? I believe uniformly equivalent can be replaced with equivalent, that is, for each $n \geqslant 0$, $\lVert \cdot \rVert_n$ is equivalent to the norm $\lVert \cdot \rVert$.
I do not see in the proof of his Lemma where he uses this hypothesis and here's my version of his proof without appealing to the uniform equivalence:
For each $n \geqslant 0$, let us denote by $\mathscr{H}_n$, the set $\mathscr{H}$ with the norm $\lVert \cdot \rVert_n$. Since for each $n \geqslant 0$, we have $\lVert \cdot \rVert _n$ is equivalent to $\lVert \cdot \rVert$, the embedding $i_n : \mathscr{H}_n \to \mathscr{H}$ given by $i_{n} (x) = x$ has a bounded inverse.
Thus, for each $n \geqslant 0$ and $x \in \mathscr{H}$, \begin{equation*} \lVert x \rVert _n ^2 = \lVert i_n ^{-1} x \rVert _n ^2 = \langle (i_n ^{-1} )^{*} i_n ^{-1}x , x \rangle \end{equation*} where $\langle \cdot , \cdot \rangle$ corresponds to the inner product induced from $\lVert \cdot \rVert$. Consequently, from the hypothesis, in the lemma, we have that \begin{equation*} \left\lvert\left\langle \left( ( i_n ^{-1} )^{*} i_n ^{-1} - (i_0 ^{-1} )^{*} i_0 ^{-1} \right) x , x \right\rangle \right\rvert \le \varepsilon_n \lVert x \rVert ^2, \qquad n \ge 0, x \in \mathscr{H} \end{equation*}
Hence, we have that $\lVert (i_n ^{-1} )^{*} i_n ^{-1} - (i_0 ^{-1} )^{*} i_0 ^{-1} \rVert \leqslant \varepsilon_n$ for each $n \geqslant 0$ (this follows from the fact that if $A$ is self adjoint operator then $\lVert A \rVert = \sup_{\lVert x \rVert \le 1} \lvert \langle Ax, x \rangle \rvert$). Hence $(i_n ^{-1} )^{*} i_n ^{-1} \to (i_0 ^{-1} )^{*} i_0 ^{-1}$ in the operator norm. Consequently, $i_n i_n ^* \to i_0 i_0 ^*$ since $\cdot ^{-1}$ is a continuous map from the group of invertible bounded linear operators to itself.
Now, since $k^n (\cdot, \lambda) = i_n ^* k(\cdot , \lambda)$ and $k^n (z,\lambda) = \langle i_n i_n ^{*} k (\cdot , \lambda) , k(\cdot , z) \rangle$, by taking limits, the proof is complete.
Is Shimorin assuming more than what is required? An example showing that uniform equivalence is unnecessary would be nice but I have not thought about one at this point of writing the question.
