Understanding the idea of the proof in a question.

Question

Here is the question:

Unbounded linear functional maps every open ball to $\mathbb{R}$?

Here is my questions to the answers there:

1-I am not getting the total idea of the solution and what are we exactly doing, could anyone explain to me the total idea?

2- In this paragraph : "Any other ball $B$ in $X$ is of the form $x+\delta X_1$ for some $\delta>0$. Given such $\delta$, for any fixed $t\in\mathbb R$ by the previous paragraph there exists $y\in X_1$ with $f(y)=(t-f(x))/\delta$. Then $f(x+\delta y)=t$. So $f(B)=\mathbb R$." in the first solution, I am not sure exactly why we are sure from this "there exists $y\in X_1$ with $f(y)=(t-f(x))/\delta$." could anyone explain this for me please?

3- I am not sure why $X_1=-X_1$? in the first solution, could anyone explain this for me please?

4- I do not understand why it may be necessarily to replace $x_{n}$ with $-x_{n}$?

5- why we are taking $t\in [0,1]$ and how this shows that $f(X_{1})$ contains the whole segment $[0, f(x_{n})]$ and why we want to show this?

6-convex set means the line segment between any 2 points in the set should lie totally inside the set, how is this used here?

Answer 1

(1)

• $X_1$ contains $0$, and $f(0) = 0$.
• Because $f$ is unbounded, $X_1$ must have points $x$ for which $f(x)$ is as large as desired. I.e., for any real number $r > 0$, we can find an $x$ such that $f(x) > r$.
• $X_1$ also contains every point of the line segment connecting $0$ to $x$
• Since $f$ is linear, $f(X_1)$ must likewise contain every number between $0 = f(0)$ and $f(x)$.
• Since $0 < r < f(x), f(X_1)$ must contain $r$. Since $r$ was arbitrary, every positive real number is in $f(X_1)$.
• For $r < 0$, there is some $y \in X_1$ with $f(y) = -r$, and therefore $f(-y) = r$. Since $-y \in X_1$, every negative real number is also in $f(X_1)$, as is $0$.
• Hence every real number is in $f(X_1)$.
• Every other open ball is a translation and rescaling of $X_1$. Linearity preserves translations and rescaling. But every translation and rescaling of $\Bbb R$ is still all of $\Bbb R$. So every open ball must be carried to $\Bbb R$ by $f$.

(2) $(t - f(x))/\delta$ is a real number, and he just showed that every real number is in $f(X_1)$.

(3) $\|x\| = \|-x\|$.

(4) Because until this point, we only know that $|f(x_n)| > n$. It could be either $f(x_n) > n$ or $f(x_n) < -n$. But if $f(x_n) < -n$, then $f(-x_n) = -f(x_n) > n$. So if we replace the original $x_n$ with its opposite, we can assure that $f(x_n) > n$ always.

(5) If $0 \le r \le n$, let $t = r/f(x_n)$. Because $f(x_n) > n, 0 \le t < 1$. But $f(tx_n) = tf(x_n) = r$. Thus every number $r \in [0,n]$ is the image of $f(tx_n)$ for some $t \in [0,1]$. As for why we want show this. It is because the entire point is the show that $f(X_1)$ contains all real numbers, not just arbitrarily large ones.

(6) The stuff you asked about in (5) is a convexity argument. $f$ carries the line segment from $0$ to $x$ in $X_1$ onto the line segment $[f(0), f(x)]$ in $\Bbb R$. Because of the properties of linearity and balls, it was not necessary to directly cast it in terms of convexity, but it is that convexity that makes the proof work.