Riesz's Lemma, which is 2.5-4 in Erwin Kreyszig's Introductory Functional Analysis With Applications, is as follows:

Let $Y$ and $Z$ be subspaces of a normed space $X$ (of any dimension), and suppose that $Y$ is closed and is a proper subset of $Z$. Then for every real number $\delta \in (0,1)$, there is an element $z\in Z$ such that $\Vert z \Vert =1$ and $\Vert z-y \Vert \geq \delta$ for all $y \in Y$.

Now Problem 7 in the problem set immediately following Section 2.5 in Kreyszig is as follows:

If $\dim Y < \infty$ in Riesz's Lemma, show that one can even choose $\delta = 1$.

I've read and I think I've understood fully the proof of the Lemma with $\delta \in (0,1)$, and I'm unable to figure out to modify that particular proof to include the case when $\delta = 1$. Nor am I able to give an independent proof of the assertion made in Problem 7.

Can anybody please help?

  • $\begingroup$ For future reference, the author's name is spelled Erwin Kreyszig. $\endgroup$ – epimorphic Dec 6 '14 at 14:51

Suppose $Y$ is finite dimensional. As $Y$ is a proper subspace of $Z$ you can find $z_0 \in Z$ such that $z_0 \notin Y$. Consider $Y' = Y + \langle z_o\rangle$. The dimension of $Y'$ is exactly one more than the dimension of $Y$ so $Y'$ is also finite dimensional. Consider $Y'$ as a normed space with the norm inherited from $X$.

Now you can find $z$ in $Y'$ such that $\|z\| = 1$ and $\| z-y \| = 1$ for all $y \in Y$ using that $Y'$ is finite dimensional. You can proceed like this:

  • take $z_n$ in $Y'$ with $\| z_n-y \| \geq 1-\frac{1}{n}$ for all $y \in Y$ and $\|z_n\|=1$
  • every finite dimensional normed space is a Banach space and its unit ball is a compact set in it
  • using the previous point, extract a convergent subsequence $(z_{n_k})_k$ from $(z_n)_n$
  • the subsequence converges to some $z$ in $Y'$ with $\| z-y \| \geq 1$ for all $y \in Y$ and $\|z\|=1$

Observe that $z \in Y' = Y + \langle z_o\rangle \subseteq Z$ and $\| z-y \| \geq 1$ for all $y \in Y$. Also, $\|z\|=1$ so $dist(z,Y) =inf \{ \|z-y\|:y\in Y \}= 1$.

  • $\begingroup$ thank you very much for answering my question so beautifully. However, there's one point which I would like to raise. After you've shown that $\Vert z \Vert=1$ and $\Vert z - y \Vert \geq 1$ for all $y \in Y$, in the very last sentence of your answer, you've stated that since $\Vert z \Vert = 1$ also, so $\Vert z-y \Vert = 1$. How did you conclude this last equality? $\endgroup$ – Saaqib Mahmood Dec 6 '14 at 14:02
  • $\begingroup$ Typo corrected ;) $\endgroup$ – Pipicito Dec 6 '14 at 14:17
  • 1
    $\begingroup$ Just in case, the infimum is $1$ because $\| z - 0\| = \| z \| = 1$ and $0 \in Y$ $\endgroup$ – Pipicito Dec 6 '14 at 14:19
  • $\begingroup$ Wonderful! Excellent!! $\endgroup$ – Saaqib Mahmood Dec 6 '14 at 14:53
  • $\begingroup$ how do you conclude that $\dist(z,Y) = 1$ from what has gone before? $\endgroup$ – Saaqib Mahmood Feb 11 '15 at 17:51

If $Y$ is finite dimensional, we also consider $Z$ to be finite dimensional (by throwing away some parts of $Z$ if necessary). Then for each $\delta = 1-\frac{1}{n}$, apply Riesz lemma so that there is $z_n \in Z$, $||z_n = 1||$ and

$$||z_n - y|| \geq 1-\frac{1}{n}$$

for all $y\in Y$. Then as $Z$ is finite dimensional, there is $z\in Z$ so that $z_n \to z$. Thus $||z||=1$ and

$$||z- y|| \geq 1$$

for all $y\in Y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.