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Which of the following statements are true and why?

  1. Any continuous function from the open unit interval $(0, 1)$ to itself has a fixed point.

  2. $\log x$ is uniformly continuous on $( 1/2,+\infty)$.

  3. If $A, B$ are closed subsets of $[0,\infty)\,$, then $A + B = \{x + y\; |\; x \in A,\, y \in B\}$ is closed in $[0,\infty)$

  4. A bounded continuous function on $\mathbb{R}$ is uniformly continuous.

  5. Suppose $f_n(x)$ is a sequence of continuous functions on the closed interval $[0, 1]$ converging to $0$ pointwise. Then the integral $\int_0^1f_n(x)\mathrm dx\,$ converges to $0$.

My thoughts:

  1. I am not sure asthe interval is not closed.
  2. It is true as it has bounded derivative.
  3. Usually, the sum of two closed set is not closed but is the case here?
  4. Not sure.
  5. Not sure.

Can anyone help me please to solve the problems? Thank you.

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  1. $x^{2}$ is good too.
  2. You are right.
  3. Let $z_{n}\in A+B$ such that $z_{n}\rightarrow z$. So $z_{n}=a_{n}+b_{n}$ with $a_{n}\in A$ and $b_{n}\in B$. If $b_{n}$ or $a_{n}$ is unlimited we get the absurd (note that $z_{n}$ converges) of $a_{n}+b_{n}$ being unlimited, because $b_{n}\geq 0$ and $a_{n}\geq 0$. So $a_{n}$ and $b_{n}$ are limiteds and then you can extract subsequences that are convergents. Now you can conclude.
  4. Take the function $$x \mapsto \sin(x^2)$$ This function is continuous, limited, but not uniformly continuous. Can u see it?
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  1. Define $f(x)=\sqrt{x}$, for every $x \in (0,1)$. Then ...

  2. Join the point $(0,n)$ and the point $(1/n,0)$ with a segment, then join $(1/n,0)$ and $(1,0)$ with a horizontal segment. This is the graph of a function $f_n$ such that $f_n \to 0$ pointwise and $\int_0^1 f_n(x)\, dx = 1/2$ for each $n$.

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  • 2
    $\begingroup$ The problem of your $f_n$ is that $f_n(0)=n$. But we can take a quite similar function, taking the line $(0,0)$ to $(n^{-1},0)$, $(n^{-1},0),(\frac{3n}2,n)$, $(\frac{3n}2,n),(2/n,0)$ and $0$ for the other points. $\endgroup$ – Davide Giraudo Sep 25 '12 at 11:47
  • $\begingroup$ @DavideGiraudo Yes, you are definitely right. The idea is to have a "triangle" of constant area that slides to a spike near some point. $\endgroup$ – Siminore Sep 25 '12 at 12:05
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  1. Or $f(x) = x^n$ or $x^{\frac{1}{n}}$.

  2. Yes. Lipschitz continuity implies uniform continuity and $|f'| = |\frac1x| \leq 2$.

  3. Yes. It is true for intervals: $+([a,b] \times [c,d]) = [a+c, b+d]$. Hence it is true for all closed sets.

  4. To add to Kaye's answer: to "break" uniform continuity (while still being continuous) you are looking for a bounded continuous function with unbounded derivative. That is, you need your function to become arbitrarily steep in places. $x \mapsto \sin x^2$ has this property (check it!).

5.Define $$ f_n(x) = \begin{cases} n^2 x - n & x \in [\frac{1}{n}, \frac{2}{n}] \\  - n^2 x + 3n & x \in [\frac{2}{n}, \frac{3}{n}] \\ \end{cases}$$ for $n \geq 3$. This looks like this: enter image description here

(drawn using slimber)

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