Riesz's lemma

hbghlyj

Wikipedia
www-users.cse.umn.edu/~garrett/m/fun/riesz_lemma.pdf
This little lemma is the Banach-space substitute for one aspect of orthogonality in Hilbert space. In a Hilbert spaces $Y$, given a non-dense subspace $X$, there is $y∈Y$ with $|y| = 1$ and $\inf_{x∈X} |x - y| = 1$, by taking $y$ in the orthogonal complement to $X$. Of course, this device is unavailable more generally.

Lemma: (Riesz) For a non-dense subspace $X$ of a Banach space $Y$, given $r<1$, there is $y \in Y$ with $|y|=1$ and $\inf _{x \in X}|x-y| \geq r$

Proof: Take $y_1$ not in the closure of $X$, and put $R=\inf _{x \in X}\left|x-y_1\right|$. Thus, $R>0$. For $\varepsilon>0$, let $x_1 \in X$ be such that $\left|x_1-y_1\right|<R+\varepsilon$. Put $y=\left(y_1-x_1\right) /\left|x_1-y_1\right|$, so $|y|=1$. And
\begin{align*}
\inf _{x \in X}|x-y|&=\inf _{x \in X}\left|x+\frac{x_1}{\left|x_1-y_1\right|}-\frac{y_1}{\left|x_1-y_1\right|}\right|\\
&=\inf _{x \in X}\left|\color{red}{\frac{x}{\left|x_1-y_1\right|}}+\frac{x_1}{\left|x_1-y_1\right|}-\frac{y_1}{\left|x_1-y_1\right|}\right|\\&=\frac{\inf _{x \in X}\left|x+x_1-y_1\right|}{\left|x_1-y_1\right|}
\\&=\frac{\inf _{x \in X}\left|\color{red}{x}-y_1\right|}{\left|x_1-y_1\right|}
\\&=\frac R{\left|x_1-y_1\right|}\\&>\frac{R}{R+\varepsilon}
\end{align*}
By choosing $\varepsilon>0$ small, $R /(R+\varepsilon)$ can be made arbitrarily close to 1 .

---

其中, 标红色的使用了$X$是子空间:
第一处红色字使用$x\in X\Leftrightarrow\frac{x}{\left|x_1-y_1\right|}\in X$
第二处红色字使用$x\in X\Leftrightarrow x+x_1\in X$

---

例子: $X=\Bbb R^2$, $Y$是$X$的子空间, 那么$Y$可以是一条通过原点的直线:

当$\varepsilon$很小时,因为$\left|x_1-y_1\right|<R+\varepsilon$,所以$x_1$和$y_1$的连线与$Y$接近垂直
再normalize得在单位圆上的就$y=\left(y_1-x_1\right) /\left|x_1-y_1\right|$,所以$y$和0的连线接近于“与$Y$垂直的那条半径”, 而半径的长度是1, 然后就不难理解$\inf _{x \in X}|x-y|$可以接近1了.

hbghlyj

A Conceptual Proof of Riesz's Lemma

Wikipedia's article on Riesz's lemma links to lecture notes by Paul Garret for a proof. Garrett presents the lemma as an independent substitute for orthogonality in Banach spaces. Unfortunately, this approach makes the proof technical and (I think) unmotivated.

There's a very clear proof that reduces the problem to the other substitute for orthogonality in Banach spaces: Hahn-Banach. The downsides are twofold. First, Hahn-Banach uses the axiom of choice (well, the Boolean Prime Ideal Theorem) in an essential way. If you're doing choiceless functional analysis (but who does that?), then Garrett's proof is your only option. Second, the clear proof doesn't seem to be easily available on the web. I can't do much about the first problem, but I can fix the second!

To show:

For $X$ a normed vector space, $Y$ a closed proper subspace and $\epsilon\in[0,1]$, there exists $x\in X$ of unit norm such that, for all $y\in Y$, $$|x-y|\geq1-\epsilon$$

Proof:

Fix any point $z\notin Y$, so that $d(z,Y)\geq 0$. Note that $d(\cdot,Y)$ is

• a sublinear function on $X$ that
• vanishes on $Y$ and
• equals "projection onto $\mathbb{R}z\cong\mathbb{R}$" on $Y+\mathbb{R}z$.

(This is not enough to prove the theorem, since we have no control over $|z|$.)

By Hahn-Banach, the linear portion of $d(\cdot,Y)$ extends to a linear functional on all of $X$ that is also dominated by $d(\cdot,Y)$. Thus $Y$ is contained in the kernel of some nonzero functional $f$.

Without loss of generality, we may rescale $f$ to have unit norm, and and take some "approximate normer" $x\in X$ such that $|f(x)|\geq1-\epsilon$. But now, for any $y\in Y$, $$(1-\epsilon)-0=|f(x)-f(y)|\leq|f||x-y|=|x-y|$$

QED.

hbghlyj

注意2#的$X,Y$和1#相反
MathOnline在Functional Analysis有很多页面, 包括Riesz's lemma

Theorem 1 (Riesz's Lemma): Let $(X, \| \cdot \|)$ be a normed linear space and let $Y \subseteq X$ be a proper and closed linear subspace of $X$. Then for all $\epsilon$ such that $0<\epsilon<1$ there exists an element $x_0 \in X$ with $\| x_0 \| = 1$ such that $\| x_0 - y \| \geq 1 - \epsilon$ for every $y \in Y$.