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The following is a homework problem (I have solved the majority of it, but need help with the last part)

Suppose $f$ : $V \rightarrow \mathbb{F}$ is a non-zero continuous linear functional on a Banach space $V$ over $\mathbb{F}$.

I. Show that W = $\{x \in V : f(x) = 1\}$ is non-empty, closed and convex.

II. Show that $$\inf_{x \in W} \|x\| = \frac{1}{\|f\|}$$ III. Show that if the set $ Y = \{x \in V : \|x\| = 1\}$ is compact then there exists $x_0 \in W$ such that $\|x_0\|$ is equal to the infimum in part II.

I'm not sure how to do part III. My teacher suggested the following: consider a sequence ${x_n}$ in $W$ such that $\|x_n\|$ is decreasing to $\displaystyle \inf_{x \in W} \|x\|$. Then $\frac{x_n}{\|x_n\|}$ is a sequence in $Y$. Since $Y$ is compact, this sequence has some subsequence converging in $Y$, i.e. $$\frac{x_{n_k}}{\|x_{n_k}\|} \rightarrow y \in Y$$ Then since $f$ is continuous: $$f\left(\frac{x_{n_k}}{\|x_{n_k}\|}\right) \rightarrow f(y)$$ Using the fact that $f$ is linear and that $f(x_{n_k}) = 1$, $$\frac{1}{\|x_{n_k}\|} \rightarrow f(y)$$

The sequence $\frac{1}{\|x_{n_k}\|}$ is increasing to $\displaystyle \frac{1}{\inf_{x \in W} \|x\|}$, so $f(y) = \displaystyle \frac{1}{\inf_{x \in W} \|x\|}$.

I am not sure how to proceed from here. Ultimately I have to show that $\exists x_o \in W$ such that $\displaystyle \|x_0\| = \inf_{x \in W} \|x\|$, but I can't see where this comes from.

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Hint (For brevity I denote the $\inf_W \|x\|$ by $c$): You have $f(y) = 1/c$ and $\|y\| = 1$. What can you say about $cy$? –  martini May 15 '12 at 15:09
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So $\|cy\| = c$ and $f(cy) = cf(y) = 1$. That means $cy \in W$ with $\|cy\| = c$, i.e. $x_0 = cy$. Thanks! –  rt93 May 15 '12 at 15:40
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1 Answer

Try this:

By definition of $||f||$, there exists $x_n$ of unit norm so that $|f(x_n)| \rightarrow ||f||$. By the compactness assumption, there is a $\hat{x}$ and a subsequence so that $x_{n_k} \rightarrow \hat{x}$ (of unit norm, of course). Then let $\hat{y} = \frac{\hat{x}}{||f||}$. Note that $\hat{y} \in W$, and $||\hat{y}|| = \frac{1}{||f||}$, as desired.

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