7
$\begingroup$

There are a couple of questions on this site regarding this problem, but I just can't seem to figure this out.

Here is the question:

Let $f$ be a continuous function on $\mathbb{R}$. A point $x$ is called a shadow point of $f$ if there is a number $y > x$ with (f(y) > f(x)). The rationale for this terminology is indicated in Figure 9; the parallel lines are the rays of the sun rising in the east (you are facing north). Suppose that all points of ((a,b)) are shadow points, but that $a$ and $b$ are not shadow points.
Clearly, $f(a) \ge f(b)$.

(a) Suppose that $f(a) > f(b)$. Show that the point where $f$ takes on its maximum value on $[a,b]$ must be $a$.

(b) Then show that this leads to a contradiction, so that in fact we must have $f(a) = f(b)$.

This little result, known as the Rising Sun Lemma, is instrumental in proving several beautiful theorems that do not appear in this book.

My proof for (a): If $f(a)$ were not the maximum point on $[a,b]$, then $f(x)>f(a)$ for some $x$ on $(a,b)$. But then $a$ would be a shadow point, a contradiction. I think this proof is correct, but I keep getting lost on part (b). I know that by continuity of $f$ at $a$ and the fact that $f(a)>f(b)$ there exist $x$ in some neighborhood of a with $f(x)>f(b)$. I just can't seem to use that to make a contradiction.

$\endgroup$
3
  • $\begingroup$ You’re on the right track.. compactness and contradiction may be your friend here.. assume they aren’t equal.. your argument for (a) works just fine.. $\endgroup$ Commented 18 hours ago
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. $\endgroup$ Commented 17 hours ago
  • $\begingroup$ Run ≠ Sun...... $\endgroup$ Commented 6 hours ago

2 Answers 2

Reset to default
1
$\begingroup$

Here's another full proof that uses the Intermediate Value Theorem. Probably not the one the book is intending, though.

Lemma: If $x \geq b$ then $f(x) \leq f(b)$. Proof: $b$ is not a shadow point.

Proof: Now, choose a value $c$ such that $f(b)<c<f(a)$.

Let $S = \{x \in [a,b] : f(x) = c\}$, which is closed as a preimage, and bounded. So $S$ is compact. It is also nonempty by IVT, so $x_0 = \max S$ exists.

I claim $x_0$ is not a shadow point. If it were, then there would be a $y > x_0$ such that $f(y) > c > f(b)$. By the lemma, $y < b$. So again applying the intermediate value theorem, there is a $z \in (y,b)$ such that $f(z) = c$. But then $z \in S$ and $z > \max S$, a contradiction.

$\endgroup$
1
$\begingroup$

Here is a low tech argument. It seems like not what the book intends, since I don't really need part (a).

Towards a contradiction, suppose that $f(a) > f(b)$. Pick some $r > 0$ so that $f(a) - r > f(b)$. Consider the set $X\subseteq (a,b]$ of points $x\in (a,b)$ so that $f(x) > f(a) - r$. By continuity of $f$ (near $a$), $X$ is non-empty so that $s = \sup X$ is defined. If $s < b$, then $s\in (a,b)$, so $s$ is a shadow point. This implies there is a $t > s$ such that $f(t) > f(s) > f(b)$. Because $b$ is not a shadow point, for $y > b$, $f(b) \ge f(y)$, so we must have $s < t < b$, so that $t\in X$. The resulting contradiction implies that $s = b$. By continuity of $f$ (near $b$), we get $f(b) \ge f(a) - r > f(b)$. This final contradiction implies that $f(a) = f(b)$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.