Banach-valued power series are determined by their real values

For every real $h$ with $\mathrm{\mid }h\mathrm{\mid }<r$ the series ${\sum }_{n\ge 0}{d}_n{h}^n$ converges in $Y$ and has sum $0$: at the point $z=z_0+h$ both given series converge, and the partial sums of the difference series are the differences of the corresponding partial sums of the two given series, so they converge to the difference of the two sums, which the hypothesis makes $0$.

Continuity at the centre. Let $(c_j{)}_{j\ge 0}\subseteq Y$ and $\rho >0$ be such that ${\sum }_{j\ge 0}{c}_j{h}^j$ converges for every real $h$ with $\mathrm{\mid }h\mathrm{\mid }<\rho$. Then its sum $S(h)$ satisfies $S(h)\to c_0$ as $h\to 0$. Indeed, put $q:=\rho \mathrm{/}2$. Convergence at $h=q$ makes the partial sums Cauchy, so their successive differences $c_j{q}^j$ tend to $0$; a sequence in a normed space that tends to $0$ is bounded, so there is $M<\mathrm{\infty }$ with $\mathrm{\parallel }{c}_j\mathrm{\parallel }{q}^j\le M$ for all $j$. For $\mathrm{\mid }h\mathrm{\mid }\le q\mathrm{/}2$ and every $N$ the tail bound $\mathrm{\parallel }{\sum }_{j>N}{c}_j{h}^j\mathrm{\parallel }\le {\sum }_{j>N}\mathrm{\parallel }{c}_j\mathrm{\parallel }\mathrm{\mid }h{\mathrm{\mid }}^j\le M{\sum }_{j>N}{2}^{-j}$ holds: the tail is the limit of its partial sums, the norm is continuous by [L2], and each partial sum is estimated by the triangle inequality. The finite part ${\sum }_{j\le N}{c}_j{h}^j$ tends to $c_0$ as $h\to 0$, and $S(0)=c_0$. Hence for $\epsilon >0$ one chooses $N$ with $M{\sum }_{j>N}{2}^{-j}<\epsilon \mathrm{/}2$ and then $h$ so small that $\mathrm{\parallel }{\sum }_{j\le N}{c}_j{h}^j-c_0\mathrm{\parallel }<\epsilon \mathrm{/}2$, giving $\mathrm{\parallel }S(h)-c_0\mathrm{\parallel }<\epsilon$.

For every $n\ge 0$: if $d_0=\cdots =d_{n-1}=0$, then $d_n=0$. Indeed, the series $T_n(h):={\sum }_{j\ge 0}{d}_{n+j}{h}^j$ converges at $h=0$ with sum $d_n$, and for $0<\mathrm{\mid }h\mathrm{\mid }<r$ the vanishing of the initial coefficients makes the partial sums of ${\sum }_{k\ge 0}{d}_k{h}^k$ equal to $h^n$ times the partial sums of $T_n(h)$, so that ${\sum }_{k\ge 0}{d}_k{h}^k=h^n{T}_n(h)$; by [step 1.1] the left side converges to $0$, hence $T_n(h)=0$ for $0<\mathrm{\mid }h\mathrm{\mid }<r$ and the series $T_n(h)$ converges for every real $\mathrm{\mid }h\mathrm{\mid }<r$. Applying [step 1.2] with $c_j:=d_{n+j}$ and any $\rho \in (0,r)$ gives $d_n=T_n(0)={\lim }_{h\to 0}{T}_n(h)=0$.

Induction on $n$: [step 2.1] says that the vanishing of $d_0,\dots ,d_{n-1}$ forces the vanishing of $d_n$ for every $n$, so the set of indices with $d_n=0$ contains $0$ and is closed under successors; it is therefore all of $\mathbb{N}$. Hence $a_n=b_n$ for every $n$.

For every complex $z$ with $\mathrm{\mid }z-z_0\mathrm{\mid }<r$ the two series are termwise identical, hence, both being convergent there, they have the same sum; this proves the agreement on the whole disc, and the argument used only limits, norm estimates and induction, so no choice principle was used.