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Let $L$ be a continuous linear functional on a Banach space $X$, put $Y=\ker L$. Suppose there is a nonzero $x_0$ in $X$ such that $\|x_0+y\|\ge\|x_0\|$ for all $y\in Y$. Prove that $\|L\|=|L(x)|$ for some $x$ in the closed unit ball.

This was an old functional analysis qualifying exam problem. The original statement asked to prove an iff statement, and the converse was easy. This direction however I'm pretty clueless about. Does anybody have a hint?

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    Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – cpiegore Aug 18 '23 at 15:18
  • You can apply the following result to $x := \frac{x_0}{|x_0|}$: Distance between point and linear Space – Bruno B Aug 18 '23 at 15:23
  • @BrunoB Very cool solution, thank you. Would you post it as an answer so I can accept it? –  Aug 18 '23 at 15:41

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Let $x := \frac{x_0}{\|x_0\|}$. Using the fact that, by assumption: $1 = \|x\| = \min_{y \in \ker(L)} \|x + y\|$ (by dividing the $y$s in the assumption by $\|x_0\|$ and relabeling them), and that in general $\inf_{y \in \ker(L)} \|x + y\| = \frac{|L(x)|}{\|L\|}$, we obtain: $1 = \frac{|L(x)|}{\|L\|}$, hence: $|L(x)| = \|L\|$.

Bruno B
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