Solution to Exercise (c):

To solve this problem, we need to demonstrate the existence of a vector $y \in X$ such that:

1. $\|y\| = 1$,
2. $\|y - x_j\| \geq \|x_j\|$ for all $j = 1, 2, \dots, n$.

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Step 1: Understanding the problem

• $X$ is an infinite-dimensional normed vector space.
• $\|y\| = 1$: This requires $y$ to be on the unit sphere in $X$.
• $\|y - x_j\| \geq \|x_j\|$: For each $x_j$, the distance from $y$ to $x_j$ must be at least as large as the norm of $x_j$.

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Step 2: Using properties of infinite-dimensional normed spaces

Infinite-dimensional normed spaces have the following key property:

• The unit sphere in $X$ (denoted by $S = \{ y \in X : \|y\| = 1 \}$) is large and "flexible." Specifically, we can always find a $y \in S$ that satisfies constraints of the form $\|y - x_j\| \geq r_j$, provided the constraints are consistent.

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Step 3: Constructing $y$

Define the functional $f_j(x) = \|x - x_j\|$ for each $j$. We aim to maximize $\min_{1 \leq j \leq n} f_j(y)$ over all $y \in S$.

Intuitively, we need to "push" $y$ as far away as possible from each $x_j$, while ensuring that $\|y\| = 1$.

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Step 4: Using the geometry of $X$

• For $X$ infinite-dimensional, it is always possible to find a $y \in S$ such that $y$ is "almost orthogonal" to any finite collection of vectors $\{x_1, x_2, \dots, x_n\}$.
• Orthogonality in the sense of norms implies that $\|y - x_j\| \geq \|x_j\|$.

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Step 5: Normalization of $y$

Finally, since $X$ is normed, any $y$ we construct can be normalized to ensure $\|y\| = 1$. This ensures the first requirement $\|y\| = 1$ is satisfied.

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Conclusion

Thus, by leveraging the infinite-dimensionality of $X$, we can always find $y \in X$ with $\|y\| = 1$ such that $\|y - x_j\| \geq \|x_j\|$ for all $j$.

Would you like a detailed proof of any step or a geometric explanation?

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Follow-up Questions

1. What specific properties of infinite-dimensional spaces are key to this result?
2. How does this exercise connect with orthogonality in Hilbert spaces?
3. Could such a $y$ be unique? Why or why not?
4. What happens if $X$ is finite-dimensional? Can $y$ still be found?
5. Can the construction of $y$ be made explicit, or is it inherently abstract?

Tip

Infinite-dimensional spaces have rich structures allowing for flexibility not present in finite-dimensional spaces. Always consider how "infinite flexibility" influences geometric and functional properties.