Continuity of the distance from one dimensional subspaces in Banach spaces

Question

Let $X$ be a Banach space, call $S_X$ its unit sphere and let $x\in X$ be fixed. Is it true that the function $f:S_X\rightarrow\mathbb{R}$ given by $y\mapsto\Vert x+\langle y\rangle\Vert$ (i.e. the distance of $x$ from the subspace spanned by $y$) is continuous? So far I only managed to prove its upper semicontinuity. If $\{y_n\} \subset S_X$ is a sequence of vectors converging to some $y\in S_X$, observe that for each $n$ and each scalar $t$ one has $\Vert x+\langle y_n\rangle\Vert\leq\Vert x-ty_n\Vert$. Passing to the limit on both sides one gets: $$ \limsup_{n\to+\infty}\Vert x+\langle y_n\rangle\Vert\leq\Vert x-ty\Vert $$ Finally, taking the infimum on the right side over all scalars $t$ leads to the formula: $$ \limsup_{n\to+\infty}\Vert x+\langle y_n\rangle\Vert\leq\Vert x+\langle y\rangle \Vert $$ Which proves the upper semicontinuity. What about lower semicontinuity? Any suggestion or help is much appreciated.

Answer 1

Fix a sequence $y_n \in S_X$ such that $y_n \to y$. First, by the reverse triangle inequality, we have that $$\|x + \lambda y_n\| - \|x+ \lambda y\| \leq |\lambda| \|y_n - y\|.$$ For fixed $\varepsilon > 0$ pick $\lambda_0$ such that $\|x + \lambda_0 y\| \leq \|x + \langle y \rangle \| + \frac{\varepsilon}{2}$. Then we get that $$\|x + \langle y_n \rangle \| \leq \|x + \lambda_0 y_n \| \leq |\lambda_0| \|y_n - y\| + \|x + \lambda_0 y\| \leq |\lambda_0| \|y_n - y\| + \|x + \langle y \rangle \| + \frac{\varepsilon}{2}$$ Then, for sufficiently large $n$, $\|y_n - y\| \leq \frac{\varepsilon}{2(1+|\lambda_0|)}$ and hence $\|x + \langle y_n \rangle \| - \|x + \langle y \rangle \| \leq \varepsilon$.

It remains to to see that for sufficiently large $n$, $\|x + \langle y \rangle \| - \|x + \langle y_n \rangle \| \leq \varepsilon$ also. This is where I will need the assumption that $\|y_n\|$ is bounded away from $0$.

Here, we pick a sequence $\lambda_n$ such that $\|x + \lambda_n y_n\| \leq \|x + \langle y_n \rangle \| + \frac{\varepsilon}{2}$. Since $y_n \in S_X$, we have that $$|\lambda_n| \leq \|x + \lambda_n y_n\| + \|x\| \leq \|x + \langle y_n \rangle \| + \frac{\varepsilon}{2} + \|x\| \leq \|x + y_n\| + \frac{\varepsilon}{2} + \|x\| \leq 1 + \frac{\varepsilon}{2} + 2\|x\|.$$

Hence $\lambda_n$ is a bounded sequence. The result now follows by the same technique as we used for the first case. Indeed, by the reverse triangle inequality, we have for $\lambda \in \mathbb{R}$ $$\| x + \lambda y \| - \|x + \lambda y_n\| \leq |\lambda| \| y_n - y\|$$

As a result, we see that $$\| x + \langle y \rangle \| \leq \|x + \lambda_n y\| \leq |\lambda_n| \|y_n - y\| + \|x + \lambda_n y_n\| \leq |\lambda_n| \|y_n - y\| + \|x + \langle y_n \rangle \| + \frac{\varepsilon}{2}$$

Now, since $\lambda_n$ is bounded, for sufficiently large $n$ we have that $|\lambda_n| \|y_n - y\| \leq \frac{\varepsilon}{2}$ and so $\| x + \langle y \rangle \| - \|x + \langle y_n \rangle\| \leq \varepsilon$.

In total, this proves that for sufficiently large $n$, $|\|x + \langle y \rangle\| - \|x + \langle y_n \rangle \|| \leq \varepsilon$ which is the desired result.