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Suppose that we have a function $f(x)$ that is continuous and bounded on $(0,1)$. Let us then define a function $g(x)=x(1-x)f(x)$. Prove that g(x) is continuous. I solved it through $\delta,\varepsilon$ definition of uniform continuity. However, I didn't need to use the fact that $f(x)$ was continuous; I basically implied through my proof that for $g(x)$ to be uniformly continuous, $f(x)$ only needed to be bounded. Which I think is wrong. Here is what I tried $$\vert g(x)-g(x_0)\vert$$$$\Rightarrow\vert x(1-x)f(x)-x_0(1-x_0)f(x_0)\vert$$$$\Rightarrow\vert x(1-x)f(x)-x(1-x)f(x_0)+x(1-x)f(x_0)-x_0(1-x_0)f(x_0)\vert$$$$\leq\vert x(1-x)\vert\vert f(x)-f(x_0)\vert+\vert f(x_0)\vert\vert x(1-x)-x_0(1-x_0)\vert$$

Since $f(x)$ is bounded we have that $\vert f(x)\vert\leq M$ for all x with M being some number. Thus we have that $$\vert x(1-x)\vert\vert f(x)-f(x_0)\vert+\vert f(x_0)\vert\vert x(1-x)-x_0(1-x_0)\vert$$$$\leq2M\vert x(1-x)\vert+M\vert x(1-x)-x_0(1-x_0)\vert$$ Afterwards, I used calculus to find the maximum of $x(1-x)$. I took derivative and set it equal to 0. This yielded me that $x(1-x)$ maximum is at $x=1/2$. Hence, $$2M\vert x(1-x)\vert+M\vert x(1-x)-x_0(1-x_0)\vert$$$$<2M\vert1/4\vert+M\vert1/4\vert=3M/4$$ Hence $\delta=4\varepsilon/3M$. Since $\delta$ does not depend on $x_0$, $g(x)$ is uniformly continuous. What did I do wrong? What are other methods to prove this?

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  • $\begingroup$ I do not see how you go from $3M/4$ to $\delta < 4\epsilon/3M$. $\endgroup$
    – user99914
    Oct 27, 2013 at 19:57
  • $\begingroup$ It should be an equal sign, sorry. If the question is more general: I thought it was similar as to how $\epsilon$ proofs for limits of sequences. Just that in this case, we are trying to express $\delta$. We were given a function, and found an expression, $\delta$, as to how far apart $x$ and $x_0$ are. We later related this to how far apart the $f(x)$ and $f(x_0)$ would be given delta. We want that $f(x)$ and $f(x_0)$ be within $\epsilon$. I am using the method presented in Ross and Lebl. $\endgroup$ Oct 27, 2013 at 20:23
  • $\begingroup$ How do you show that $|g(x) - g(x_0)|\leq \epsilon$ for your choice of $\delta$? $\endgroup$
    – user99914
    Oct 27, 2013 at 20:34
  • $\begingroup$ We would expand |g(x)−g(x0)|≤ϵ and express the $\vert x-x_0\vert$ as $\delta$. Plug in the expression for $\delta$ and see whether |g(x)−g(x0)|≤ϵ holds. I am not quite sure what you're trying to get at. Could you elaborate? $\endgroup$ Oct 27, 2013 at 20:55
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    $\begingroup$ Dont you come up with $|g(x) - g(x_0)| < 3M/4$? How do you come up with $\epsilon$? $\endgroup$
    – user99914
    Oct 27, 2013 at 22:16

2 Answers 2

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Hint: Let $\varepsilon > 0$ be given. Assume that $|f(x)| \le M$. Since $f$ is uniformly continuous on $\left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$ you can choose $\tilde\delta$ such that $|f(x)-f(y)| < \varepsilon$ when $x,y \in \left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$ and $|x-y| < \tilde\delta$. Now put $\delta := \min\left\{\tilde\delta, \dfrac{\varepsilon}{4M}\right\}$.

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  • $\begingroup$ How did you determine the interval? I understand that a continuous function is uniformly continuous on a closed interval but I am not quite sure why you chose $\varepsilon/4M$ $\endgroup$ Oct 27, 2013 at 21:05
  • $\begingroup$ If $x<\dfrac{\varepsilon}{4M}$ and $|x-y|<\dfrac{\varepsilon}{4M}$, then $y<\dfrac{\varepsilon}{2M}$ and hence $|x(1-x)f(x)| \le \dfrac{\varepsilon}{2}$, $|y(1-y)f(y)| \le \dfrac{\varepsilon}{2}$. Similarly consider other cases. $\endgroup$
    – njguliyev
    Oct 27, 2013 at 21:13
  • $\begingroup$ I am confused, aren't we saying that x and y are elements of the closed set? And wouldn't that imply that both of them cannot be smaller than $\varepsilon/4M$? This is assuming that $\varepsilon/4M$ is smaller than 1, of course. But I cannot see how it would work even if $\varepsilon/4M$ is greater than 1. $\endgroup$ Oct 27, 2013 at 21:24
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    $\begingroup$ For given $\varepsilon$ we chose some $\delta$. Now we have to prove that for any $x,y \in (0,1)$ with $|x-y|< \delta$ we get $|f(x)-f(y)| < \varepsilon$. If both of these points $\in \left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$, then there is nothing to check. Otherwise, at least one of $x,y$ is outside that closed interval. Then try to consider all possible cases using my hints. $\endgroup$
    – njguliyev
    Oct 27, 2013 at 21:31
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Hi‎nt: If ‎$‎‎f(x)$ ‎is ‎continuous ‎in ‎‎$‎‎(a,b)$ ‎and ‎there ‎are‎ $‎‎lim‎_{x\to a^+}f(x), ‎‎lim‎_{x\to b^-}f(x)‎$, then ‎$‎‎f(x)$ ‎is‎ ‎uniformly ‎continuous.

‎Since ‎$‎x(1-x)‎$ ‎and ‎‎$‎f(x)‎$ ‎are ‎continuous ‎in‎ $‎‎(0,1)$‎, ‎then ‎‎$‎‎g(x)$ ‎is ‎continuous ‎in‎ $‎‎(0,1)$. Also, since there are ‎$‎‎lim‎_{x\to 0^+}g(x)‎$ ‎and‎‎ ‎$‎‎lim‎_{x\to 1^-}g(x)‎$‏, ‎so ‎‎$‎‎g(x)$ ‎is ‎uniformly ‎continuous.‎

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