Diagonalization over a normal function and its derivatives on transfinite ordinals

Question

Let $\Phi(0,\beta)$ a normal function from $On$ to $On$, and let $\Phi(\alpha,\beta)$ be the $\alpha$-th derivative of $\Phi(0,\beta)$. For example, let $\Phi(0,\beta)=\aleph_\beta$. Then, all functions $F_\alpha(\beta)=\Phi(\alpha,\beta)$ are also normal functions for a fixed $\alpha$. I have two questions:

1. Let $G_\beta(\alpha)=\Phi(\alpha,\beta)$. Is this one also a normal function for fixed $\beta$? (Maybe it depends on $\beta$?)
2. Is the diagonalization $\Psi(\alpha)=\Phi(\alpha,\alpha)$ also a normal function?

Answer 1

1. No, not all $G_\beta$ are normal. For example let $\Phi(0,\beta)$ be any normal function whose least fixed point is greater than $\omega$ and consider $G_\omega(\alpha)$. Since $\Phi(\beta+1,0)>\omega$ we have $G_\omega(k)<\Phi(k+1,0)$, and by increasingness we also have $\Phi(k,0)<G_\omega(k)$ for each natural $k$. We get the following chain: $$\Phi(1,0)<G_\omega(1)<\Phi(2,0)<G_\omega(2)<\Phi(3,0)<G_\omega(3)<\ldots$$ By definition $\Phi(\omega,0)=\textrm{sup}\{\Phi(k,0)\mid k<\omega\}$, and the ordinals $G_\omega(k)$ are unbounded in the set of $\Phi(k,0)$, so $\textrm{sup}\{G_\omega(k)\mid k<\omega\}=\Phi(\omega,0)<G_\omega(\omega)$.
2. No, since the above point shows discontinuity of $\Psi$: $\textrm{sup}\{\Psi(k)\mid k<\omega\}=\Phi(\omega,0)<\Psi(\omega)$.

Answer 2

At least you have that $\Phi(\alpha,0)$ is indeed normal. It is increasing, because $\Phi(\alpha+1,0)$ is always greater than $\Phi(\alpha,0)$; and it is also continuous since, by definition, $\Phi(\lambda,0) = sup\{\Phi(\beta,0)|\beta<\lambda\}$ for limit ordinal $\lambda$.