这是一份BATH巴斯大学MA10274作业代写的成功案例
First we show that $|y|$ is necessarily 1 . In fact, if we let $c=(1-|y|) / 2$. then $0 \leq c \leq 1$. Considering the vectors
$$
x_{1}=\frac{(1-c) x+c y}{|y|} \text { and } y_{1}=\frac{(1-c)(y)+c x}{|y|}
$$
we find that $x_{1}, y_{1} \in B_{X}$ and $\left|x_{1}-y_{1}\right|=\varepsilon$. Therefore, .
Rut
$$
\left|x_{1}+y_{1}\right|=\frac{1}{|y|}|\dot{x}+y|_{.} .
$$
Since $|y| \leq 1$, it follows from this last inequality and our choice of $x, y$ that il. $\left.y\right|^{\prime}=1$.
Having ascertained that $|y|=1$, what about $x$ ? Of course, if $|x|=1$, too. then we are done. Suppose $|x|<1$. Pick $\varphi \in S_{X}$ * so that
$$
\varphi\left(\frac{x+y}{|x+y|}\right)=1 \text {. }
$$
MA10274 COURSE NOTES ：
$$
\left(2\left[1-\delta_{\left|x_{a} \ln z\right|}(k)\right]\right)^{p}<2 $$ Since given $x, y \in S_{X}$ for which $|x-y| \geq k$ we have $$ |x+y| \leq 2\left(1-\delta_{\left\{x_{n} \ln _{A 21}\right.}(k)\right) . $$ it follows that the continuous functions $\varphi(t)$ and $\chi(t)$ given by $$ \varphi(t)=|x+t y|^{p}, \quad \chi(t)=1+t^{p} $$ satisfy $\varphi(1)<\chi(1)$; consequently, there is an $\eta>0$ so that
$$
|x+t y|^{p} \leq 1+t^{p},
$$
whenever $|1-t| \leq \eta$. Of course, we can assume $\eta$ is very small, say $\eta<1$.
Claim. For any finitely nonzero sequence $\left(a_{m}\right)$ of scalars we have
$$
\frac{\eta}{2}\left|\sum_{m} a_{m} x_{m}\right| \leq\left|\left(a_{m}\right)\right|_{p} .
$$