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If F Uniformly Continuous on Two Intervals

Written By mercredi 19 octobre 2022 Add Comment Edit

  • Is it true that if $S$ is an unbounded set of real numbers and MATH for every number $x\in S$ then the function $f$ fails to be uniformly continuous?
    The assertion given here is false. Every function defined on the set $Z$ of integers must be uniformly continuous.

  • Given that MATH prove that $f$ is continuous but not uniformly continuous on the set MATH .
    Since the number $2$ does not belong to the domain of $f$ , the function $f$ is constant in a neighborhood of every number in its domain. Therefore $f$ is continuous on the set MATH . To see why $f$ fails to be uniformly continuous we observe that given any positive number $\delta $ we can find a number MATH and a number MATH such that MATH , and for any such choice of numbers $x$ and $t$ we must have MATH

  • Given that MATH for all real numbers $x,$ prove that $f$ is not uniformly continuous on the set $R$ .
    Figure
    We define MATH and MATH for every positive integer $n$ . Since MATH and MATH for every $n$ we know that MATH does not approach $0$ as MATH . Now MATH as MATH and so it follows from the relationship between limits of sequences and uniform continuity that the function $f$ fails to be uniformly continuous.

  • sn-logo.wmf Ask Scientific Notebook to make some 2D plots of the function $f$ defined by the equation MATH for $x>0$ . Plot the function on each of the intervals MATH , MATH , MATH and MATH . Revise your plot and increase its sample size if it appears to contain errors. Why do these graphs suggest that $f$ fails to be unformly continuous on the interval MATH ? Prove that this function does, indeed, fail to be uniformly continuous.

    Solution: To prove that $f$ fails to be uniformly continuous we shall show that for every number $\delta >0$ there exist two positive numbers $a$ and $b$ such that MATH and MATH . We begin by choosing a number $p$ such that whenever $x\geq p$ we have MATH Now choose a positive integer $n$ such that MATH Since $x\log x>n\pi $ for $x$ sufficiently large $x$ , we can use the Bolzano intermediate value theorem to choose a number $a>p$ such that $a\log a=n\pi $ . Now since MATH we can use the Bolzano intermediate value theorem again to choose a number MATH such that MATH We now observe that MATH and so the proof is complete.

    1. A function $f$ is said to be Lipschitzian on a set $S$ if there exists a number $k$ such that the inequality MATH holds for all numbers $t$ and $x$ in $S$ . Prove that every Lipschitzian function is uniformly continuous.
      Suppose that $f$ is a function defined on a set $S$ , that $k$ is a positive number and that the inequality MATH holds for all numbers $t$ and $x$ in the set $S$ . Suppose that $\varepsilon >0$ . We define MATH and observe that, whenever $t$ and $x$ belong to $S$ and MATH we have MATH

    2. Given that MATH for all MATH prove that $f$ is uniformly continuous but not lipschitzian on $\left[ 0,1\right] $ .

      Solution: The fact that $f$ is uniformly continuous on the closed bounded set $\left[ 0,1\right] $ follows at once from the fact that $f$ is continuous there. Now, to prove that $f$ fails to be Lipschitzian, suppose that $k$ is any positive number. Given MATH we see that MATH and this exceeds $k$ whenever $x<1/k^{2}$ .

    1. Suppose that $\ f$ is uniformly continuous on a set $S,$ that MATH is a sequence in the set $S$ and that MATH has a partial limit $x\in R.$ Prove that it is impossible to have MATH as MATH .
      Using the fact that $f$ is uniformly continuous on $S$ we choose a number $\delta >0$ such that the inequality MATH holds whenever $t$ and $s$ belong to $S$ and MATH . Since $x$ is a partial limit of the sequence MATH we know that the condition MATH holds for infinitely many positive integers $n$ . Choose a positive integer $N $ such that MATH For every one of the infinitely many positive integers $n$ for which the condition MATH holds, since MATH we have MATH Therefore the sequence of numbers MATH is frequently in the interval MATH and so it cannot approach $\infty $ .
      Of course, this sequence cannot approach $-\infty $ either.

    2. Did you assume that $x\in S$ in Part a? If you did, go back and do the exercise again. You have no information that $x\in S$ . If you didn't assume $x\in S,$ you can sit this question out.

    3. Suppose that $f$ is uniformly continuous on a bounded set $S$ and that MATH is a sequence in $S.$ Prove that it is impossible to have MATH as MATH .
      The assertion follows at once from Part a and the fact that every bounded sequence of numbers has a partial limit in $R$ .

    4. Prove that if $f$ is uniformly continuous on a bounded set $S$ then the function $f$ is bounded.

      Solution: To obtain a contradiction, assume that $f$ is unbounded above. Choose a sequence MATH in $S$ such that MATH for each $n$ . Now use part c of this exercise.
      Well, of course, such a choice of MATH would make MATH which we know now to be impossible.

    1. Given that $S$ is a set of real numbers, that MATH and that MATH for all $x\in S$ , prove that $f$ is continuous on $S$ but not uniformly continuous.
      Hint: Use the preceding exercise. Choose a sequence MATH in MATH that converges to $x$ . Note that MATH as MATH .

    2. Given that $S$ is a set of real numbers and that $S$ fails to be closed, prove that there exists a continuous function on $S$ that fails to be uniformly continuous on $S$ .
      Choose a number MATH and use Part a.

    3. Is it true that if $S$ is an unbounded set of real numbers then there exists a continuous function on $S$ that fails to be uniformly continuous on $S$ ?
      No, as we remarked after Exercise 1, every function on the set $Z$ of integers must be uniformly continuous.

  • Is it true that the composition of a uniformly continuous function with a uniformly continuous function is uniformly continuous?
    Yes, this assertion is true. Suppose that $f$ is a uniformly continuous function on a set $S$ and that $g$ is a uniformly continuous function on a set $T$ that includes the range $f\left[ S\right] $ of $f$ . To show that the function $g\circ f$ is uniformly continuous on $S$ , suppose that $\varepsilon >0$ .
    Using the uniform continuity of $g$ on the set $T$ we choose a number $\delta >0$ such that the inequality MATH holds whenever $u$ and $v$ belong to $T$ and MATH . Now, using the uniform continuity of $f$ on the set $S$ we choose a positive number $\gamma $ such that the inequality MATH will hold whenever $s$ and $t$ belong to $S$ and MATH . Then whenever $s$ and $t$ belong to $S$ and MATH we have MATH .

    1. Suppose that $f$ is uniformly continuous on a set $S$ and that MATH is a convergent sequence in $S$ . Prove that the sequence MATH cannot have more than one partial limit.
      We know from the result proved in Exercise 6 d that the sequence MATH is bounded. We choose a partial limit $y$ of MATH and we want to prove that $y$ is the only partial limit of MATH . Suppose that $z $ is any number unequal to $y$ . We define MATH Choose a number $\delta >0$ such that the inequality MATH holds whenever $s$ and $t$ belong to $S$ and MATH .
      We write the limit of MATH as $x$ and choose an integer $N$ such that the inequality MATH holds whenever $n\geq N$ . Using the fact that $y$ is a partial limit of the sequence MATH we choose an integer $m\geq N $ such that MATH . Now given any integer $n\geq N$ , since MATH we must have MATH and consequently MATH
      Figure
      Thus if $n\geq N$ , the number MATH cannot lie in the neighborhood MATH of $z$ and we conclude that $z$ fails to be a partial limit of the sequence MATH .

    2. In Part a, did you assume that the limit of the sequence MATH belongs to $S$ ? If so, go back and do the problem again.

    3. Prove that if $f$ is uniformly continuous on a set $S$ and MATH is a convergent sequence in $S$ then the sequence MATH is also convergent. Do not assume that the limit of MATH belongs to $S$ .
      In view of Part a, the present result follows at once from an earlier theorem on limits of sequences.

    4. Suppose that $f$ is uniformly continuous on a set $S,$ that $x$ is a real number and that MATH and MATH are sequences in $S$ that converge to the number $x.$ Prove that MATH The existence of these limits was guaranteed in Part c. Now since MATH as MATH we deduce from the relationship between uniform continuity and limits of sequences that MATH as MATH .

    5. Suppose that $f$ is uniformly continuous on a set $S$ and that MATH . Explain how we can use Part d to extend the definition of the function $f$ to the number $x$ in such a way that $f$ is continuous on the set MATH .
      We know that there exists a sequence MATH in $S$ that converges to $x$ and we know that there is a single limit for all of the sequences MATH that can be made in this way. We define $f\left( x\right) $ to be this common limit value. This extension of the function $f$ to the set MATH is uniformly continuous. The proof will be given in the more extended case that we consider below in Part f.

    6. Prove that if $f$ is uniformly continuous on a set $S$ then it is possible to extend $f$ to the closure $\overline{S}$ of $S$ in such a way that $f$ is uniformly continuous on $\overline{S}$ .
      For every number MATH we define $f\left( x\right) $ by the method described in Part e. To show that the extension of $f$ to $\overline{S}$ is uniformly continuous, suppose that $\varepsilon >0$ . Using the uniform continuity of $f$ on $S$ we choose $\delta >0$ such that the inequality MATH holds whenever $t$ and $x$ belong to $S$ and MATH . We shall now observe that whenever $t$ and $x$ belong to $\overline{S}$ and MATH we must have MATH . To make this observation, suppose that $t$ and $x$ belong to $\overline{S}$ and that MATH .
      choose a sequence MATH in $S$ that converges to $t$ and a sequence MATH in $S$ that converges to $x$ . Since MATH we know that the inequality MATH must hold for all $n$ sufficiently large and therefore, since MATH and since MATH for all $n$ sufficiently large we must have MATH

  • Suppose that $f$ is a continuous function on a bounded set $S$ . rove that the following two conditions are equivalent:

    1. The function $f$ is uniformly continuous on $S$ .

    2. It is possible to extend $f$ to a continuous function on the set $\overline{S}$ .

    The fact that condition a implies condition b follows from Exercise 9. On the other hand, if $f$ has a continuous extension to the set $\overline{S}$ then, this extension, being a continuous function on a closed bounded set, must be uniformly continuous; and so $f$ must be uniformly continuous on $S$ .

  • Given that $f$ is a function defined on a set $S$ of real numbers, prove that the following conditions are equivalent:

    1. The function $f$ fails to be uniformly continuous on the set $S$ .

    2. There exists a number $\varepsilon >0$ and there exist two sequences MATH and MATH in $S$ such that MATH as MATH and MATH for every $n$ .

    At the suggestion of my good friend Sean Ellermeyer this exercise was upgraded to a theorem. I have left in the exercise. Sometimes I find it interesting to see which of my students recognize that an item is the same as one they have already seen.

    • rileytakentlyr.blogspot.com

    Source: https://sites.radford.edu/~wyang/430/page216/Uniform%20Continuity-hint.htm

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