I studying Brezis' book and I have somes partial solutions of the exercise $1.17.$

Let be $E$ a normed space and $f\in E^*$ be a linear functional nonzero. Consider the set $M=[f=0]$ given by

$$M=\{x\in E \ : \ \langle f,x\rangle = 0 \}.$$

Here, $\langle f,x\rangle$ denote $f(x)$ and note that $M$ is just $\ker f$.

Dertemine $M^\perp.$

By definition

$$M^\perp=\{g\in E^* \ : \ \langle g,x\rangle=0, \forall x\in M \}.$$

So, the first doubt is: $M^\perp$ is just

$$M^\perp=\{g\in E^* \ : \ \ker f \subset \ker g \} ?$$

Show that $\forall x\in E$, we have $d(x,M)=\inf_{y\in M}{\|x-y\|}=\frac{|\langle f,x\rangle|}{\|f\|}.$

Since $y\in M$, $|\langle f,x\rangle|=|\langle f,x-y\rangle|\le \|f\|\|x-y\|$. Then

$$\frac{|\langle f,x\rangle|}{\|f\|}\le d(x,M).$$

How can I show the equality? I know that since $\{x\}$ is a compact set and $M$ is a closed subspace of $E$, there is $y_0\in M$, such that $d(x,M)=\inf_{y\in M}{\|x-y\|}=\|x-y_0\|,$ but I don't know if that it is a help.

I tried to show that $\forall \varepsilon >0$, there is $z\in M$ such that

$$\|x-z\|<\frac{|\langle f,x\rangle|}{\|f\|}+\varepsilon,$$

but I could not. The last doubt is:

Let be $E=\{u\in C([0,1],\Bbb{R}) \ : \ u(0)=0\}$ and $f\in E^*$ given for each $u\in E$ by $$\langle f,u\rangle = \int_0^1 u(t)dt.$$ Show that if $u\in E-M$ then $d(u,M)$ is never achieved.

For that, I tried something by absurd method but I found nothing. Can someone give a little help? Thanks.


2 Answers 2


You have $M=\mathcal{N}(f)$. Then $g\in M^{\perp}$ iff $\mathcal{N}(f)\subseteq\mathcal{N}(g)$. Suppose $f(u)\ne 0$ for some $u\in X$.Then $f\left(y-\frac{f(y)}{f(u)}u\right)=0$ holds for all $y\in X$, which gives $$ 0= g\left(y-\frac{f(y)}{f(u)}u\right)=g(y)-\frac{f(y)}{f(u)}g(u),\;\;\;y\in X \\ \implies g = \frac{g(u)}{f(u)}f. $$ Therefore $M^{\perp}$ is the linear subspace of $E^{\star}$ spanned by $f$. And, \begin{align} \mbox{dist}(x,M) & =\sup_{g\in M^{\perp},\|g\|=1}|g(x)| \\ & = \sup_{g\in M^{\perp},g\ne 0}\frac{|g(x)|}{\|g\|} = \frac{|f(x)|}{\|f\|}. \end{align} For the last part, please explain your use of $x$ and $u$; I think there must be a typo.

Addressing your question in the comments: If $M^c$ is the closure of $M$, then \begin{align} \mbox{dist}(x,M) & =\mbox{dist}(x,M^c) \\ & = \|x+M^c\|_{X/M^c} \\ & = \sup_{y^{\star}\in (X/M^c)^{\star},\|y^{\star}\|=1}|y^{\star}(x+M^c)| \\ & = \sup_{g\in X^{\star}\cap M^{\perp},\|g\|=1}|g(x)| \\ & = \sup_{g\in X^{\star}\cap M^{\perp},g\ne 0}\frac{|g(x)|}{\|g\|} \end{align} If $\mbox{dist}(x,M)\ne 0$, then $x\notin M$, which means $f(x)\ne 0$. Therefore, $g \in M^{\perp}$ is $g=\frac{g(x)}{f(x)}f$ as explained above, where it was shown that an equality holds for all $u$ for which $f(u)\ne 0$.

  • $\begingroup$ Thanks for your help, I'll explain the last part here, but before why we have that $d(x,M)=\sup_{g\in M^\perp, \|g\|=1}|g(x)|$? I don't see that. $\endgroup$
    – Irddo
    Apr 2, 2016 at 14:51
  • $\begingroup$ @Irddo : I added a little at the end. $\endgroup$ Apr 2, 2016 at 15:36
  • $\begingroup$ Sorry for the delay. I really studied your solution (the second part) and I don't understood many things. So, I would like your help again. In the second equatily, why I can take the distance like $\|x+M^c\|$? Cause I don't know if the infimum is achieved. In the third equality, why I take the supremum just in $(X/M^c)^*$? Thanks again. $\endgroup$
    – Irddo
    Apr 4, 2016 at 20:26
  • $\begingroup$ @Irddo : $x+M^c$ denotes an equivalence class. It is an element of $X/M^c$. The norm of an element of $X/M^c$ is the inf of the $X$ norms of elements of the equivalence class. In this case, that inf is $\inf_{m\in M^c}\|x+m\|=\inf_{m\in M^c}\|x-m\|$, which you can see is the same as $\mbox{dist}(x,M)$. Right? $\endgroup$ Apr 4, 2016 at 22:09
  • $\begingroup$ @TheTrialAndError, sorry for my mistake, it's ok now. Thanks. So, for the third part of the problem, have you any help? I tried that: Suppose that there is $v\in M$ such that $\|u-v\|=|\int_0^1 u(t)dt|,$ for $u\in E-M.$ So $\int_0^1 u(t)-v(t)\le |\int_0^1u(t)dt|,$ then $\int_0^1u(t)-|u(t)|dt\le 0$. I stopped here. $\endgroup$
    – Irddo
    Apr 4, 2016 at 22:39

We can show the inequality $d(x,M)\leqslant\frac{f(x)}{\|f\|}$ by an alternative way. For any fixed $u\in E\setminus M$, we have that $$y_x:=x-\frac{f(x)}{f(u)}u\in M,$$ hence $$d(x,M)=\inf_{y\in M}\|x-y\|\stackrel{(y_x\in M)}\leqslant\|x-y_x\|=\left\|x-x-\frac{f(x)}{f(u)}u\right\|=\frac{|f(x)|}{|f(u)|}\|u\|,$$ which implies $$\frac{|f(u)|}{\|u\|}\leqslant\frac{|f(x)|}{d(x,M)},\quad\forall u\in E\setminus\{0\}.$$ Remark. Of course, the above inequality is obtained by taking $u\in E\setminus M$, however it still holds if $u\in M$.

It follows from the above inequality that $$\|f\|=\sup_{u\in E\setminus\{0\}}\frac{|f(u)|}{\|u\|}\leqslant\frac{|f(x)|}{d(x,M)}\quad\Longrightarrow\quad d(x,M)\leqslant\frac{|f(x)|}{\|f\|}.$$


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