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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?

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  • $\begingroup$ For future reference, the author's name is spelled Erwin Kreyszig. $\endgroup$ – epimorphic Dec 6 '14 at 14:51
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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 $dist(z,Y) =inf \{ \|z-y\|:y\in Y \}= 1$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $y\in Y$.

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