<s>[INST] If $\|x^*\|=1$ and $\|x\|=\|y\|=1$ then $\|x-y\|<\epsilon$ Suppose $X$ is a space with inner product. Let $x^{\ast} \in X^{\ast}, \| x^{\ast} \| = 1$, and $0 < \varepsilon < 1$. If
$x, y \in X, \| x \| = \| y \| = 1, x^{\ast} (x) > 1 -
\frac{\varepsilon^2}{8}$, and $x^{\ast} (y) > 1 - \frac{\varepsilon^2}{8}$,
then $\| x - y \| < \varepsilon$.
I know
$$
     | x^{\ast} (z) | \leqslant \| x^{\ast} \|  \| z \|
    \text{ for all } z \in X $$
Then
$$ | x^{\ast} (x) | \leqslant 1 \text{ and } | x^{\ast} (y) | \leqslant 1 $$
Now by hypothesis
$$ \begin{align}
     1 - \frac{\varepsilon^2}{8} &< x^{\ast} (x) \leqslant 1\\
     - 1 &\leqslant - x^{\ast} (y) < \frac{\varepsilon^2}{8} - 1\\[1em]
     \hline 
     - \frac{\varepsilon^2}{8} &< x^{\ast} (x) - x^{\ast} (y) <
     \frac{\varepsilon^2}{8}\\
     | x^{\ast} (x - y) | &< \frac{\varepsilon^2}{8},
   \end{align} $$
and also I have that
$$ | x^{\ast} (x - y) | \leqslant \| x - y \| . $$
But I don't know what else can I do to arrive at
$$ \| x - y \| < \varepsilon . $$
Any suggestion will be welcome.
 [/INST] By the parallelogram law we have
\begin{equation}
  \| x + y \|^2 + \| x - y \|^2 = 2 (\| x \|^2 + \| y \|^2) \implies \| x +
  y \|^2 = 4 - \| x - y \|^2 \tag{1}
\end{equation}
Also by hypothesis
\begin{align}
     x^{\ast} (x) &> 1 - \frac{\varepsilon^2}{8}\\
     x^{\ast} (y) &> 1 - \frac{\varepsilon^2}{8}\\  
\\ \hline
     \| x^{\ast} \|  \| x + y \| &\geqslant x^{\ast} (x + y) > 2 -
     \frac{\varepsilon^2}{4}\\
     \Rightarrow \text{  } \| x + y \| &> 2 - \frac{\varepsilon^2}{4}
   \end{align}
Using the last result and (1), we have
\begin{align}
     4 - \| x - y \|^2 &> \left( 2 - \frac{\varepsilon^2}{4} \right)^2\\
     4 - \| x - y \|^2 &> 4 - \varepsilon^2 + \frac{\varepsilon^4}{16}\\
     \| x - y \|^2 &< \varepsilon^2 - \frac{\varepsilon^4}{16} <
     \varepsilon^2\\
     \| x - y \| &< \varepsilon .
   \end{align}
</s>