Norm on a Banach space

Question

Let $\left( X, \| \cdot \|\right) $ be a Banach space over some field $\mathbb{K}$.

Let $x$ be fixed in $X$ such that $\|x\| \le 1$.

If $x_0$ is any point in $X$ , I need to show that there exists $\alpha $ in $\mathbb{K}$ such that $\| x+ \alpha x_0\| =1$.

Could anyone guide me as to how I could prove the existence of $\alpha$?

Thanks

Answer 1

If $\|x\|=1$, it is trivial, else prove that $\exists \beta\in\mathbb{K}$ such that $\|x+\beta x_0\|>1$ (use triangular inequality) and use the fact that the norm is continuous.

Answer 2

Hint: The function $f \colon \alpha \mapsto \|x+\alpha x_0\|$ is continuous on $[0, +\infty)$, $f(0) \le 1$, and $f(\alpha) = |\alpha|\|x_0 + \frac1\alpha x\| \to +\infty$, $\alpha \to +\infty$ if $\|x_0\| \ne 0$.

Answer 3

If $ \|x\|=1$ choose $\alpha=1$. Otherwise define $g(\alpha):=\|x+\alpha x_0\|$. Then $g$ is continuous, $g(0)<1$. Note that $x_0\neq0$ and verify using the triangle inequality that $$g\left(\frac{2-\|x||}{\|x_0\|}\right)\geq2.$$

Michael