The question
I have a family of Reproducing Kernel Hilbert Spaces $\mathcal{B}$. A Reproducing Kernel Hilbert Space (RKHS) $H_k$ is a Hilbert Space together with a positive semi-definite function $k$ of which every function $f\in H_k$ can be written like $f(x)=\int k(y,x)f(y)dy$. This function $k$ uniquely defines the RKHS. Consider the kernel $k_\pi$ given by \begin{align*} k_\pi(x,y) := \mathbb{E}_{w\sim\pi}[\sigma(w^Tx)\sigma(w^Ty)] = \int \sigma(w^Tx)\sigma(w^Ty)\pi(dw) \end{align*} where $\pi$ is some probability distribution and $\sigma(x) = \max(0,x)$. If we take the union over all the possible probability distribution $\pi$, then we get \begin{align*} \mathcal{B} := \cup_\pi H_{k_\pi} \end{align*} This union is indicated by the letter $\mathcal{B}$, since it is commonly called the Barron Space.
Because of the specific choices for $k_\pi$ every function in $H_{k_\pi}$ can be written like \begin{align*} f(x) = \int a(w)\sigma(w^Tx)\pi(dw) \end{align*} for some $a$. As a result we have that the norms of $H_{k_\pi}$ and $\mathcal{B}$ are given by \begin{align*} ||f||_{H_{k_\pi}} &= \int |a(w)|^2\pi(dw) \\ ||f||_{\mathcal{B}} &= \inf_\pi ||f||_{H_{k_\pi}} \\ \end{align*}
I was wondering whether $\mathcal{B}$ is again a Hilbert Space. Since we have a norm, the simplest way to check this would be to check the parallelogram law. This states that if a space $X$ has a norm satisfying \begin{equation} ||x+y||^2+||x-y||^2 \leq 2||x||^2+2||y||^2 \quad \forall x,y\in X \end{equation} So does $||\cdot||_\mathcal{B}$ satisfy the parallelogram law?
... and my attempt at solving it
If both $f,g\in H_{k_{\pi_i}}$ for some $\pi_i$, then both $f,g\in\mathcal{B}$ and \begin{align*} ||f+g||_\mathcal{B} &= ||f+g||_{H_{k_{\pi_i}}} \\ ||f-g||_\mathcal{B} &= ||f-g||_{H_{k_{\pi_i}}} \\ ||f||_\mathcal{B} &= ||f||_{H_{k_{\pi_i}}} \\ ||g||_\mathcal{B} &= ||g||_{H_{k_{\pi_i}}} \\ \end{align*} hence the parallellogram law is satisfied by virtue of $H_{k_{\pi_i}}$ being a Hilbert Space.
Now w.l.o.g. let $f\in H_{k_{\pi_0}},g\in H_{k_{\pi_1}}$ with $\pi_0$ and $\pi_1$ distinct. Then we know that \begin{align*} ||f+g||_\mathcal{B} &= \inf_\pi ||f+g||_{H_{k_{\pi}}} \leq ||f+g||_{H_{k_{\pi_j}}} \quad j\in\{0,1\} \\ ||f-g||_\mathcal{B} &= \inf_\pi ||f-g||_{H_{k_{\pi}}} \leq ||f-g||_{H_{k_{\pi_j}}} \quad j\in\{0,1\} \\ \end{align*} Now we can write \begin{align*} ||f+g||^2_\mathcal{B} + ||f-g||^2_\mathcal{B} &\leq ||f-g||^2_{H_{k_{\pi_j}}}+||f-g||^2_{H_{k_{\pi_j}}} \\ &= 2||f||^2_{H_{k_{\pi_j}}}+2||g||^2_{H_{k_{\pi_j}}} \\ \end{align*} for $j\in\{0,1\}$. Taking $j=0$ gives \begin{align*} ||f+g||^2_\mathcal{B} + ||f-g||^2_\mathcal{B} &\leq 2||f||^2_{H_{k_{\pi_0}}}+2||g||^2_{H_{k_{\pi_0}}} \\ &= 2||f||^2_\mathcal{B}+2||g||^2_{H_{k_{\pi_0}}} \\ &\color{red}{\geq 2||f||^2_\mathcal{B}+2||g||^2_\mathcal{B}} \end{align*} whereas taking $j=1$ gives \begin{align*} ||f+g||^2_\mathcal{B} + ||f-g||^2_\mathcal{B} &\leq 2||f||^2_{H_{k_{\pi_0}}}+2||g||^2_{H_{k_{\pi_0}}} \\ &= 2||f||^2_{H_{k_{\pi_1}}}+2||g||^2_\mathcal{B} \\ &\color{red}{\geq 2||f||^2_\mathcal{B}+2||g||^2_\mathcal{B}} \end{align*} Both lines in red contradict the parallellogram law. Does this mean that $\mathcal{B}$ is no Hilbert Space? Or am I missing some step in my proof?