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.
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}