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This is an exercise from my lecture on functional analysis. Let $$E := \{ u ∈ C^0 ([0, 1]) : u(0) = 0 \}, \quad \|·\| := \|·\|_\infty$$ and $$\ f : E \longrightarrow\mathbb{R}, \quad f(u) := \int_{0}^{1} u(t) dt $$ Let $M$ be the set $$M := \{u ∈ E : f(u) = 0 \}$$ I was able to prove that $\|f\|_{E'}=1$ and $\text{dist}(u,M)=|f(u)|$. I have to prove that for given $u \in E \setminus M$, there is no $u' \in M$ such that $\text{dist}(u, M) = \|u − u'\|_\infty$.

My attempt

By contradiction if exists $u'$ such that $\|u − u'\|_∞= \left|\int_{0}^{1} u(t)dt\right|$ then I want to show that $u(x)=u'(x)$ for every $x \in [0,1]$, that is of course a contradiction. Any idea or hint to conclude?

Thank you for the help.

user773458
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oel
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  • Since we're working with continuous functions, $u(x)=u'(x)$ for every $x\in[0,1]$ is equivalent to $|u-u'|_{\infty}=0$. If $u\notin M$, then $|f(u)|\neq 0$, so I don't think you'll be able to prove the contradiction you want. Can you think of any properties that $M$ as a set could have that would guarantee that the distance from $u'\in M$ to the complement of $M$ is an infimum and not a minimum? – Otto Baier Oct 10 '24 at 17:01
  • Actually, are you sure that $u\in E\setminus M$ and $u'\in M$ and not the other way around? I believe that the statement "for every $u\in E\setminus M$, there exists a $u'\in M$ such that $\text{dist}(u,M)=|u-u'|_{\infty}$" is true. – Otto Baier Oct 10 '24 at 17:10
  • @OttoBaier Honestly, I don't have much faith in this demonstration strategy either, but that was the hint my professor wrote me. The statement is as it is written in my workbook, with the hint to do it by contradiction. – oel Oct 10 '24 at 17:19

1 Answers1

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Hint: Can the functional $f$ attain its norm? That is, can there exist a point $u \in E$ with norm one such that $1=\|f\|=|f(u)|$?

Spoiler: No

Can you conclude?

Evangelopoulos Foivos
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    Yes, it sufficies to set $v=\frac{u-u'}{|u-u'|}$, it follows that $v$ has norm one and by linearity $|f(v)|=1$ which is impossible. Thank you for the hint. – oel Oct 10 '24 at 17:40