Problem 1. Let E be a real vector space, F be a closed subspace of E, and a € E\F. Set d(a, F) inf{||a - r||; x € F}. 1. Show that if F ha a finite dimension then there exists zo € F such that d(x, F) = ||x-xo||.

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part 1 functionnal analysis
**Problem 1**

Let \( E \) be a real vector space, \( F \) be a closed subspace of \( E \), and \( a \in E \setminus F \). Set \( d(a, F) = \inf \{ \|a - x\|; x \in F \} \).

1. Show that if \( F \) has a finite dimension, then there exists \( x_0 \in F \) such that \( d(x, F) = \|x - x_0\| \).

2. Let \( \varphi \) be a nonzero linear continuous form on \( E \) and \( a \in E \) such that \( \varphi(a) \neq 0 \). We consider \( F = \ker \varphi \).

   (a) Show that \( \|\varphi\| = \frac{|\varphi(a)|}{d(a, \ker \varphi)} \).

   (b) Show that \( \frac{\varphi(u + ta)}{\|u + ta\|} \leq \frac{|\varphi(a)|}{d(a, \ker \varphi)} \) for all \( u \in \ker \varphi \). (Hint: Recall that \( E = \ker \varphi + \mathbb{R}a \)).

   (c) Deduce that \( \|\varphi\| = \frac{|\varphi(a)|}{d(a, \ker \varphi)} \).

   (d) Show that if \( u_0 \in \ker \varphi \) such that \( d(a, \ker \varphi) = \|a - x_0\| \) then there exists \( x_0 \in E \) such that \( \|x_0\| = 1 \) and \( |\varphi(x_0)| = \|\varphi\| \).

   (e) Let \( x_0 \in E \) so that \( x_0 = v_0 + ta \) with \( v \in \ker \varphi \). Suppose that \( \|x_0\| = 1 \) and \( |\varphi(x_0)| = \|\varphi\| \). Show that \( |t_0|d(a, \ker \
Transcribed Image Text:**Problem 1** Let \( E \) be a real vector space, \( F \) be a closed subspace of \( E \), and \( a \in E \setminus F \). Set \( d(a, F) = \inf \{ \|a - x\|; x \in F \} \). 1. Show that if \( F \) has a finite dimension, then there exists \( x_0 \in F \) such that \( d(x, F) = \|x - x_0\| \). 2. Let \( \varphi \) be a nonzero linear continuous form on \( E \) and \( a \in E \) such that \( \varphi(a) \neq 0 \). We consider \( F = \ker \varphi \). (a) Show that \( \|\varphi\| = \frac{|\varphi(a)|}{d(a, \ker \varphi)} \). (b) Show that \( \frac{\varphi(u + ta)}{\|u + ta\|} \leq \frac{|\varphi(a)|}{d(a, \ker \varphi)} \) for all \( u \in \ker \varphi \). (Hint: Recall that \( E = \ker \varphi + \mathbb{R}a \)). (c) Deduce that \( \|\varphi\| = \frac{|\varphi(a)|}{d(a, \ker \varphi)} \). (d) Show that if \( u_0 \in \ker \varphi \) such that \( d(a, \ker \varphi) = \|a - x_0\| \) then there exists \( x_0 \in E \) such that \( \|x_0\| = 1 \) and \( |\varphi(x_0)| = \|\varphi\| \). (e) Let \( x_0 \in E \) so that \( x_0 = v_0 + ta \) with \( v \in \ker \varphi \). Suppose that \( \|x_0\| = 1 \) and \( |\varphi(x_0)| = \|\varphi\| \). Show that \( |t_0|d(a, \ker \
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