Inner Product Space is Uniformly Convex

Theorem
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.

Then $V$ is uniformly convex.

Proof
Let $\norm \cdot$ be the inner product norm for $\struct {V, \innerprod \cdot \cdot}$.

Let $\epsilon > 0$.

Let $x, y \in V$ be such that:


 * $\norm x = \norm y = 1$

and:


 * $\norm {x - y} > \epsilon$

Then from the Parallelogram Law (Inner Product Space), we have:


 * $\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2} = 4$

So that:


 * $\norm {x + y}^2 = 4 - \norm {x - y}^2$

Since:


 * $\norm {x - y} > \epsilon$

we have:


 * $\norm {x + y}^2 < 4 - \epsilon^2$

From positive homogeneity, we have:


 * $\ds \norm {\frac {x + y} 2}^2 < 1 - \frac {\epsilon^2} 4$

We now pick $\delta$ such that:


 * $\ds \paren {1 - \delta}^2 = 1 - \frac {\epsilon^2} 4$

Note that since:


 * $\norm {x - y} \le \norm x + \norm y$

we have:


 * $\norm {x - y} \le 2$

So if there exists $x, y \in V$ with $\norm x = \norm y = 1$ such that:


 * $\norm {x - y} > \epsilon$

we have $0 < \epsilon < 2$.

Hence, if $\epsilon \ge 2$, the demand is vacuously true, and we can pick any positive real $\delta > 0$.

Now take $0 < \epsilon < 2$.

We then have:


 * $\ds 0 < 4 - \epsilon^2 < 4$

so that:


 * $\ds 0 < 1 - \frac {\epsilon^2} 4 < 1$

So, we may pick:


 * $\ds \delta = 1 - \sqrt {1 - \frac {\epsilon^2} 4} > 0$

Then, for each $x, y \in V$ such that $\norm x = \norm y = 1$ and $\norm {x - y} > \epsilon$, we have:


 * $\ds \norm {\frac {x + y} 2}^2 < \paren {1 - \delta}^2$

Since $0 < 1 - \delta < 1$, we have:


 * $\ds \norm {\frac {x + y} 2} < 1 - \delta$

Since $\epsilon > 0$ was arbitrary, for each $\epsilon > 0$ we have found $\delta > 0$ such that:


 * whenever $x, y \in V$ have $\norm x = \norm y = 1$ and $\norm {x - y} > \epsilon$, we have:
 * $\ds \norm {\frac {x + y} 2} < 1 - \delta$

as required.

So $V$ is uniformly convex.