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--- math104-f21:hw11 [2021/11/08 07:35]
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 . Is there a monotone function on $[0,1]$, such that it is discontinuous at $1/2, 1/3, 1/4, \cdots$? If so, give a construction; if no, give a proof.
+====== Solution ======
+. Assume $f: X \to Y$ is uniformly continuous, and $g: Y \to Z$ is uniformly continuous. Is it true that $g \circ f: X \to Z$ is uniformly continuous?
+Yes, it is uniformly continuous. For any $\epsilon>0$, by uniform continuity of $g$, we can find $\delta_1$, such that for $y_1, y_2$ with distance less than $\delta_1$, we have $d_Z(g(y_1), g(y_2))<\epsilon$. Using uniform continuity of $f$, we can find $\delta_2$, such that $d_X(x_1, x_2) < \delta_2$ imples $d_Y(f(x_1),f(x_2))<\delta_1$.
+. If $f: (0,1) \to (0,1)$ is a continuous map, is it true that there exists a $x \in (0,1)$, such that $f(x) = x$? What if one change the setting to $f: [0,1] \to [0,1]$ being continuous, does your conclusion change?
+For the first part of the question, the answer is no. For example, $f(x) = x^2$, then there is no $a \in (0,1)$, such that $f(a) = a$. One can see from the graph $y = f(x)$, it has no intersection with the diagonal $y=x$ over $x \in (0,1)$.
+For the second part of the question, if we are looking for $a \in (0,1)$ such that $f(a)=a$, then the answer is still no, and the counter-example above still works.
+Note that, if we are looking for $a \in [0,1]$ (closed interval), such that $f(a)=a$, then the answer is yes. Since if $f(0)=0$ or $f(1)=1$, we can take $a=0$ or $a=1$. If $f(0)>0$ and $f(1)<1$, then we can consider $g(x) = f(x)-x$, and note that $g(0)>0$, $g(1)<0$, hence by intermediate value theorem there exists $a \in (0,1)$ with $g(a)=0$, i.e. $f(a)=a$.
+. A function $f: \R \to \R$ is called a Lipschitz function, if there exists a $M>0$, such that $|f(x)-f(y)| / |x-y| \leq M$ for any $x \neq y \in \R$. Is it true that Lipschitz functions are uniformly continuous? Is it true that all uniformly continuous functions are Lipschitz?
+Yes, Lipschitz function are uniformly continuous. Since for any $\epsilon>0$, we can let $\delta = \epsilon / M$, then $|x-y| < \delta$ implies $|f(x)-f(y)| \leq M |x-y| = \epsilon$.
+Converse, it is false that all uniformly continuous functions are Lipschitz. For example, consider $f(x) = \sqrt{x}$ on $[0,1]$. It is uniformly continuous, since it is continuous on a closed interval [0,1]$. The derivative diverges at $x=0$, hence it is not Lipschitz.
+. Consider the following function $f: (0,1) \to \R$, $f(x) = x \sin (1/x)$. If $f$ uniformly continuous? Give a proof for your claim.
+Yes, it is uniformly continuous. In class we proved that $f$ can be extended to $[0,1]$, by setting $f(0)=0$. This shows the function $f$ on $[0,1]$ is uniformly continuous, hence $f$ on $(0,1)$ is uniformly continuous.
+. Is there a monotone function on $[0,1]$, such that it is discontinuous at $1/2, 1/3, 1/4, \cdots$? If so, give a construction; if no, give a proof.
+Yes, there is a monotone function. For example, $f(x) = 1/n$ for $x \in [1/(n+1), 1/n) $.

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