Let $X$ be a Banach space and $B:=(0,\varepsilon)$ be a ball. Show that $B$ and $A[B]$, where $A$ is a continuous operator, are both perfectly convex Notation: Let $X$ and $Y$ be Banach spaces and let $A\colon X\to Y$ be a bounded linear operator onto $Y$. For $\varepsilon>0$, define $B:=B(0,\varepsilon)$ to be the ball centered around zero of radius $\varepsilon$. Also needed is the following definition:

Definition: Let $X$ be a Banach space. A set $K\subseteq X$ is called perfectly convex if and only if for every bounded sequence $x_i\in K$ and for every sequence of reals $a_i\geq 0$ such that $\sum_{i=1}^{\infty}a_i=1$ we have that $\sum_{i=1}^{\infty}a_ix_i\in K$.

Currently I'm self studying functional analysis, namely perfectly convex sets in Banach spaces. In the text, the other gives the following comment in a proof:

Let $B$ be an open around zero. $B$ is perfectly convex and since the operator $A$

This isn't obvious to me, so I wrote out the proofs, but I have questions.

Proof that $B$ is perfectly convex.
Let $x_i\in B$ be a bounded sequence and $a_i\geq0$ be a sequence of reals such that $\sum_{i=1}^{\infty}a_i=1$. Since $x_i\in B$ we have $||x_i||<\varepsilon$. We want to show that $\sum_{i=1}^{\infty}a_ix_i\in B$. Notice
$$
\left|\left|\sum_{i=1}^{\infty}a_ix_i\right|\right|\leq
\sum_{i=1}^{\infty}\left|\left|a_ix_i\right|\right|=
\sum_{i=1}^{\infty}|a_i|\left|\left|x_i\right|\right|=
\sum_{i=1}^{\infty}a_i\left|\left|x_i\right|\right|<
\sum_{i=1}^{\infty}a_i\varepsilon=
\varepsilon\sum_{i=1}^{\infty}a_i=
\varepsilon\cdot1=\varepsilon.
$$
Therefore we have that $\left|\left|\sum_{i=1}^{\infty}a_ix_i\right|\right|<\varepsilon$, which implies by definition $\sum_{i=1}^{\infty}a_ix_i\in B$.
Question 1: Did we use that $x_i\in B$ were bounded here? I mean since couldn't we have just said let $x_i\in B$, and never mentioned boundedness since it somewhat comes for free by being in the set $B$?

Proof that the continuous image of $B$ is perfectly convex.
Let $\varepsilon>0$, $B:=B(0,\varepsilon)$ and consider $A[B]=\{Ax:x\in B\}$. Let $Ax_i\in A[B]$ be a bounded sequence and $a_i\geq0$ be a sequence of reals such that $\sum_{i=1}^{\infty}a_i=1$. We want to show that $\sum_{i=1}^{\infty}a_iAx_i\in A[B]$. Since $A$ is a linear operator,
$$
\sum_{i=1}^{\infty}a_iAx_i=A\left(\sum_{i=1}^{\infty}a_ix_i\right).\tag{1}
$$
Since $x_i\in B$ is a bounded sequence and $a_i\geq0$ is a sequence of reals such that $\sum_{i=1}^{\infty}a_i=1$, we have by the previous proof that $\sum_{i=1}^{\infty}a_ix_i\in B$. Therefore the expression in $(1)$ belongs to $A[B]$.
Question 2: In this proof where/why did we need that the operator $A$ is continuous and/or bounded. I suppose it is to guarantee that $x_i$ are bounded. However, as mentioned in question 1, since $x_i\in B$ aren't they automatically bounded? Hence we don't need $A$ to be bounded?
 A: The boundedness condition is here to guarantee that $\sum_{i=1}^\infty a_ix_i$ converges. Indeed, if $M>0$ is a constant such that $\|x_i\| \le M$ for all $i \in \Bbb{N}$ then
$$\sum_{i=1}^\infty\|a_ix_i\| = \sum_{i=1}^\infty a_i \|x_i\| \le \sum_{i=1}^\infty a_iM = M < +\infty$$
so the series $\sum_{i=1}^\infty a_ix_i$ converges absolutely. Since $X$ is a Banach space, it also converges in $X$.
Sure, the boundedness of your $(x_i)_{i\in\Bbb{N}}$ comes for free from being in $B$, but you need this condition in general when looking at unbounded sets.
The continuity of linear map $A$ is used to commute $A$ and the infinite sum $\sum_{i=1}^\infty$. Linearity gives that $A$ commutes with finite sums but you need continuity to commute with the limit:
$$A\left(\sum_{i=1}^\infty a_ix_i\right) = A\left(\lim_{n\to\infty}\sum_{i=1}^n a_ix_i\right) = \lim_{n\to\infty} A\left(\sum_{i=1}^n a_ix_i\right) = \lim_{n\to\infty} \sum_{i=1}^n a_iAx_i = \sum_{i=1}^\infty a_iAx_i.$$
On the other hand, $(Ax_i)_{i\in\Bbb{N}}$ bounded in general does not guarantee that $(x_i)_{i\in\Bbb{N}}$ is bounded (e.g. they can all lie in the kernel of $A$). However here it is unnecessary since $B$ is a bounded set, as you said.
