Let $x,y\in\ell^2$ satisfy $\lVert x\rVert_2=\lVert y\rVert_2=1$. The <parallelogram law> gives
$$
\left\lVert\frac{x+y}{2}\right\rVert_2^2
+\left\lVert\frac{x-y}{2}\right\rVert_2^2
=\frac{\lVert x\rVert_2^2+\lVert y\rVert_2^2}{2}=1.
$$
If $\lVert x-y\rVert_2\geq\varepsilon$, then
$$
\left\lVert\frac{x+y}{2}\right\rVert_2^2
\leq1-\frac{\varepsilon^2}{4},
$$
and therefore
$$
\left\lVert\frac{x+y}{2}\right\rVert_2
\leq\sqrt{1-\frac{\varepsilon^2}{4}}
=1-\delta,
\qquad
\boxed{\delta=1-\sqrt{1-\varepsilon^2/4}>0.}
$$
This depends only on $\varepsilon$, so the <l2 sequence space> is a <uniformly convex Banach space>, as asserted by the <uniform convexity of l2>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 21H / d / Solution

The <l2 sequence space> is a <uniformly convex Banach space>. If $\lVert x\rVert_2=\lVert y\rVert_2=1$, the <parallelogram law> gives
$$
\left\lVert\frac{x+y}{2}\right\rVert_2^2
=1-\frac14\lVert x-y\rVert_2^2.
$$
Thus $\lVert x-y\rVert_2\geq\varepsilon$ permits the modulus $\delta(\varepsilon)=1-\sqrt{1-\varepsilon^2/4}$.

Uniform convexity of l2

{title2=$\ell^2$}

The space $\ell^2$ consists of scalar sequences $x=(x_n)$ satisfying $\sum_n|x_n|^2<\infty$. With inner product $\langle x,y\rangle=\sum_nx_n\overline{y_n}$, it is a <Hilbert space>.

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 Incoming links: l2 sequence space