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Good day, We would like to know if there is a mistake in the following:

We are asked to prove the following:

We define:

$g(x)=\lim\limits_{y \to x} f(y)$

Assuming that $\lim\limits_{y \to x} f(y)$ exists for all x, prove that $g(x)$ is continuous.

I made the following prove:

We are asked to prove that: $\lim\limits_{x \to a} g(x)=g(a)$

By definition:

$g(a)=\lim\limits_{y \to a} f(y)$ so:

$$\forall \epsilon_1>0 \exists \delta_1>0 \text{ such that if } \ |y-a|<\delta_1, \text{ then } |f(y)-g(a)|<\epsilon_1$$

and

$g(x)=\lim\limits_{y \to x} f(y)$

$$\forall \epsilon_2>0 \exists \delta_2>0 | if \ |y-x|<\delta_2 \to |f(y)-g(x)|<\epsilon_2$$

if We take $$\epsilon_2=\epsilon_2=\epsilon/2$$ and let be $$\delta_3=min(\delta_1,\delta_2)$$

so:

if $$|y-a|<\delta_3 \land |y-x|<\delta_3$$ then $$|f(y)-g(a)|<\epsilon/2 \land |f(y)-g(x)|<\epsilon/2$$

but if

$$|f(y)-g(a)|<\epsilon/2 \land |f(y)-g(x)|<\epsilon/2$$ then $$|g(x)-g(a)|<\epsilon$$

but also if we have $$-\delta_3<y-a<\delta_3 \land -\delta_3<y-x<\delta_3$$ then $$-2*\delta_3<x-a<2*\delta_3$$ and if we take: $$2*\delta_3=\delta$$ then We have:

$$\forall \epsilon>0 \exists \delta>0 | if \ |x-a|<\delta \to |g(x)-g(a)|<\epsilon$$ And it is completed.

My question if there is a mistake in the prove, actually in the part where I take $$2*\delta_3=\delta$$.

thanks.

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    $\begingroup$ Are you working with functions $f: \Bbb R \to \Bbb R$? Second issue: I very much appreciate the effort you put into the question, but I'm finding the formatting makes it difficult to read. Could you please put your proposed proof into MathJax? It'll make it easier for those of us who'd like to help. $\endgroup$ Oct 6, 2020 at 3:23
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    $\begingroup$ Hi! I changed the format, I appreciate your help. And yes it is a $$f: R \to R$$ $\endgroup$ Oct 6, 2020 at 3:55

3 Answers 3

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This question is also asked here and here

That last step, of requiring $\lvert x - a \rvert < 2\delta_3$ does seem suspicious. Your proof I think requires $x$ and $a$ to be close enough together that they can both be sufficiently close to a neighborhood of y's, such that all the inequalities hold.

Be careful about adding together $\delta$ requirements, which is not always a logical thing to do. For example, If we require that $k < 1$, and also $k < 35$, adding these together doesn't really provide useful information. $2k < 36$ is still a true requirement, but it's not a sufficient one, and it obscures the more strict $k < 1$.

I suspect that requiring $\lvert x - a \rvert < \delta_3$ is sufficient. If $2 \delta_3$ works, than certainly $\delta_3$ would also. I think $2 \delta_3$ works, but just barely (I don't think it would work if $<$ was replaced with $\leq$.)

This problem comes up in Calculus by Spivak, Chapter 6, problem 16(d). I think I mostly understand the answer from Spivak. I'll try to outline the main steps, leaving details to you. Hopefully if I'm wrong about anything, someone will jump in.

Maybe this other approach will be helpful.

First, we can use the definition of limits and $g(x)$ to show that for any $\varepsilon > 0$ there is a $\delta > 0$ such that for all $y$, if $$0<\lvert y - a \rvert < \delta$$ then,

$$\lvert f(y) - g(a) \rvert < \varepsilon$$

and so,

$$g(a)-\varepsilon < f(y) < g(a) + \varepsilon$$

The next bit is tricky and I'm not sure I completely understand it:

We set $x$ within $\delta$ of $a$, and then take the limit of all sides of the above inequality for $y$ approaching $x$:

For all $x$, if

$$0<\lvert x - a \rvert < \delta$$

then,

$$\lim\limits_{y \to x} (g(a)-\varepsilon) \leq \lim\limits_{y \to x} f(y) \leq \lim\limits_{y \to x}(g(a) + \varepsilon)$$

From here, with some manipulation you should be able to show that $\lim\limits_{x \to a} g(x) = g(a).$

I'd like to return to that limit over the inequality step. First, notice that the signs change from $< to \leq$. This is a consequence of taking the limit and I won't go into the proof here, unless requested. Besides that, what's our justification for taking the limit over the inequality? How do we know that for $y$ approaching $x$, this inequality remains true? The original condition required y to be within

$$0<\lvert y - a \rvert < \delta$$

I think the idea is, because $x$ is within this same region and there is "space" around $x$, i.e. $x \neq a-\delta$, or $a+\delta$, there are open intervals around $x$ that are contained in the region. So for $y$'s very close to $x$, we know these $y$'s are in $0<\lvert y - a \rvert < \delta$, and thus the inequalities still hold.

It makes sense to pick $x$ near $a$, as we're trying to show what happens to $g(x)$ as $x$ approaches $a$.

Admittedly, it makes sense to me in retrospect. Can't say I would ever have thought to do it on my own.

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  • $\begingroup$ Hey Ben! What do you mean by "...take the limit of all sides of the above inequality for $𝑦$ approaching $𝑥$..."? $\endgroup$ Mar 30, 2021 at 21:19
  • $\begingroup$ And if you have the time, I would love to see a proof on why the signs change. $\endgroup$ Mar 30, 2021 at 21:35
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    $\begingroup$ Hi Carlevaro, this involves a few useful results from previous problems in Spivak. Problem 5-12 shows that for functions $f$ $g$ (general functions: not the $f$ and $g$ above) if $f(x) \leq g(x)$ for all $x$ around $a$, then $\lim_{x \to a} f(x) \leq \lim_{x \to a} g(x)$ provided these limits exist. Furthermore, if $f(x)$ is strictly < $g(x)$, it doesn't necessarily follow that $f(x) < g(x)$. Functions can have equal limits even when one function is always greater than the other. $\endgroup$
    – Ben
    Mar 30, 2021 at 23:23
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    $\begingroup$ For me, this problem was/is one of the most baffling. There's probably a more intuitive approach along the lines of what Robert Shore describes below. $\endgroup$
    – Ben
    Mar 31, 2021 at 0:10
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    $\begingroup$ I have to agree that is a baffling problem with a surprisingly easy solution. That's what makes it even more interesting. +1 $\endgroup$
    – Paramanand Singh
    Mar 31, 2021 at 4:37
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If I'm following your logic correctly, you're finding one neighborhood of $a$ where you can bound $\vert g(x)-f(x) \vert$ and another, necessarily overlapping neighborhood of $a$ where you can bound $\vert f(x)-g(a) \vert$. That lets you control $\vert g(x)-g(a) \vert \leq \vert g(x)-f(x) \vert + \vert f(x)-g(a) \vert$ within the overlap, which is the result you need.

I haven't chased through your arithmetic but the logic is correct.

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Sorry for the language, here is the correct demo in case it helps someone Por la hipotesis, sabemos que

$g(x)=\lim\limits_{y \to x} f(y)$

Asumimos que el $\lim\limits_{y \to x} f(y)$ existe para toda x, y debemos probar que $g(x)$ es continua.

Consideremos primero un punto particular donde hay discontinuidad evitable

$g(a)=\lim\limits_{y \to a} f(y)$ entonces:

$$\forall \epsilon_1>0 \exists \delta_1>0 \text{ such that if } \ |y-a|<\delta_1, \text{ then } |f(y)-g(a)|<\epsilon_1$$

y

$g(x)=\lim\limits_{y \to x} f(y)$

$$\forall \epsilon_2>0 \exists \delta_2>0 | if \ |y-x|<\delta_2 \to |f(y)-g(x)|<\epsilon_2$$

Si nosotros elegimos $\epsilon$ tal que $$\epsilon_1=\epsilon_2=\epsilon/2$$ and let be $$\delta<min(\delta_1,\delta_2)$$

si $$|y-a|<\delta \land |y-x|<\delta$$ then $$|f(y)-g(a)|<\epsilon/2 \land |f(y)-g(x)|<\epsilon/2$$

pero tambien

$$|f(y)-g(a)|<\epsilon/2 \land |f(y)-g(x)|<\epsilon/2$$ Sumando ambas expresiones y usando algunos propiedades y artificios del valor absoluto. $$|g(x)-g(a)|=|f(y)-g(a)-f(y) +g(x)| \leq|f(y)-g(a)|+|-f(y)+g(x)|+ |f(y)-g(a)|+|f(y)-g(x)|<\epsilon$$ Es decir \

$$|g(x)-g(a)|<\epsilon$$

$$\forall \epsilon>0 \exists \delta>0 | if \ |y-a|<\delta \to |g(x)-g(a)|<\epsilon$$ y la prueba quedaria completa.

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