I have a problem about a continuous function

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I have a problem about a continuous function

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[PREREQUISITES]
Definition of continuity and basic facts about real numbers, sequences and limits.

[GENERAL DISCUSSION]
Any discussion of how to solve problems about continuity is complicated by the fact that there are two statements that can be taken as the basic definition of continuity, one in terms of epsilons and deltas, and one in terms of sequences. (For a discussion of what they are and how the equivalence is proved, see Example 2 of [[I have a problem to solve in real analysis and I do not believe that a fundamental idea is needed]].) Sometimes, one approach is considerably slicker than the other, but often they are about equally simple and it is just a matter of taste which one prefers. We shall often just discuss one approach. There is a separate article on [[How to translate epsilon-delta proofs into sequence-based proofs and vice versa]].

===What is the problem?===

Which of the following descriptions best fits your problem?

[add] I have an explicitly defined function $f$ and I want to show that $f$ is continuous.{{
*Which of the following descriptions best fits your function $f$?
**[add]$f$ is a polynomial.{{
***In that case, one might think that the best approach was just to go ahead and prove the continuity of $f$ directly from the definition of continuity. But the calculations involved in doing this are a bit tedious and can be avoided in a number of ways. One way is a bit of a cheat: develop the theory of differentiability, prove that polynomials are differentiable, and deduce that they are continuous. A more direct approach is to prove the following four facts: (i) constant functions are continuous; (ii) the function $f(x)=x$ is continuous; (iii) if $f$ and $g$ are continuous then $f+g$ is continuous; (iv) if $f$ and $g$ are continuous then $fg$ is continuous. Any polynomial can be built up from $1$ and $x$ by means of pointwise addition and multiplication, so an easy [[induction front page|induction]] demonstrates that every polynomial is continuous.}}[/add]
**[add]$f$ is a rational function (that is, a function of the form $p(x)/q(x)$ where $p$ and $q$ are polynomials).{{
***An obvious first remark to make is that $f$ is ''not'' continuous at any $x$ where $q(x)=0.$ But if $q$ does not vanish in the range of $x$ that you are interested in, then the best approach to rational functions is very similar to what you do for polynomials. So you should read that first, and then prove one further rule (or just use it if it has been proved for you): that if $f$ and $g$ are continuous functions and $g$ does not vanish anywhere, then $f/g$ is continuous.}}[/add]
**[add]$f$ is a well-known function such as $\sin(x)$ or $e^x$.{{
***It is not quite so easy to give advice on these functions, because there are many different equivalent definitions of these functions. But if you know the addition rule for $\sin(x),$ then you can argue that $\sin(x+h)=\sin(x)\cos(h)+\cos(x)\sin(h),$ so the continuity of $\sin$ follows if you can show that $\cos(h)\rightarrow 1$ and $\sin(h)\rightarrow 0$ as $h\rightarrow 0.$ Similarly, the continuity of $e^x$ follows from the rule $e^{x+h}=e^xe^h$ and the fact that $e^h\rightarrow 1.$ But in order to prove such basic facts about $\sin$ and $\exp,$ you may well have defined these functions by means of power series and used general facts about the continuity and differentiability of power series. So a lot depends on what you are allowed to assume and how you have developed the theory.}}[/add]
**[add]$f$ is none of the above, but it is defined by a formula that involves well-known functions.{{
***The advice for polynomials and rational functions applies here. If your function is built out of functions that you know are continuous and if the method of building is one that produces continuous functions from continuous functions, then your function is continuous.}}[/add]
**[add]$f$ is not given by a straightforward formula, but rather it is defined as a limit or infinite sum of some functions $f_n.${{
***For a problem of this kind, you will want to use some theory. By far the most commonly used result for proving the continuity of a limit of functions is the following one: if the functions $f_n$ are continuous and $f_n$ [[w:Uniform convergence|converges uniformly]] to $f,$ then $f$ is continuous. If the functions $f_n$ are defined on an unbounded set such as $\R,$ then it may well be that this approach works even if they do not converge uniformly to $f,$ since often they converge ''locally uniformly''. This means that every $x$ is contained in an interval $(a,b)$ such that $f_n$ converges uniformly to $f$ on $(a,b).$ This is enough to prove that $f$ is continuous at $x,$ and since $x$ is arbitrary it proves that $f$ is continuous. This technique gives a proof that the power series $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$ converges to a continuous function, and similarly for the power series for $\sin$ and $\cos$ and a host of other functions.}}[/add]}}[/add]

[add] I have an explicitly defined function $f$ and I want to show that $f$ is not continuous.{{
*The first step, which is usually easy, is to identify a point at which you believe that $f$ is not continuous. Now your problem is that of proving that $f$ is not continuous at $x,$ for some given $x.$ Usually the quickest way of doing this is to use the sequence definition: that is, find a sequence $(x_n)$ such that $x_n\rightarrow x$ but $f(x_n)\rightarrow f(x).$ For example, let $f(x)=\sin(1/x)$ when $x\ne 0$ and $0$ when $x=0.$ Then the slickest proof that $f$ is not continuous is to note that when $x=1/(n+1/2)\pi$ then $\sin(1/x)=\sin((n+1/2)\pi)=\pm 1.$ Therefore, if we let $x_n=1/(n+1/2)\pi,$ then $x_n\rightarrow 0$ but $f(x_n)$ does not tend to $f(0)=0.$}}[/add]

[add] I have a function $f$ about which I have certain information, and I want to deduce that $f$ is continuous.{{
*At the time of writing, I do not have any good examples in mind of such problems, so the following advice may be unsatisfactory. (If it is, then please consider editing this paragraph.) However, the most obvious two methods seem to be at opposite ends of a certain spectrum: either prove the result directly from the definition of continuity, or prove it by using some theorem of the form, "Every function with such-and-such a property is continuous." Proofs in the first category may well be discoverable using the general kind of approach outlined in [[I have a problem to solve in real analysis and I do not believe that a fundamental idea is needed]]. For example, if you know that $f^{-1}(U)$ is an open set whenever $U$ is an open set and want to deduce that $f$ is continuous, then the methods in that article are sufficient. As for proofs in the second category, they may well just be extremely simple observations such as that every differentiable function is continuous.}}[/add]

[add] I have a function $f$ about which I have certain information, and I want to deduce that $f$ is not continuous.{{
*At the time of writing, I do not have any good examples in mind of such problems, so the following advice may be unsatisfactory. (If it is, then please consider editing this paragraph.) The obvious approach to a problem of this kind is to
argue directly, just as you could in the case of an explicitly defined function $f$, and use the information you have about $f$ to construct a sequence $a_n$ that converges to a limit $a$, with $f(a_n)$ not converging to $f(a)$. If such an approach doesn't work, or seems to be getting your hands unnecessarily dirty, then you could try exploiting a theorem of the form, "Every continuous function has such-and-such a property." That is, you could look for a property that your function does not have but would have to have if it were continuous. For example, suppose you know that $f:\R\rightarrow\R$ is unbounded but that $f(x)\rightarrow 0$ as $|x|\rightarrow\infty.$ Then there must be some $M$ such that $|f(x)|\leq 1$ whenever $|x|\geq M.$ Therefore, $f$ is unbounded in the interval $[-M,M].$ But continuous functions are bounded on any closed interval, so $f$ cannot be continuous.}}[/add]

[add] I have a function $f$ that I know is continuous and I want to prove something about $f$.{{
*The most obvious advice here is to use one of the following handful of basic and important results about continuous functions. (1) If $f$ is continuous and $a_n\rightarrow a$ then $f(a_n)\rightarrow f(a).$ (2) A continuous function defined on a closed bounded interval $[a,b]$ is bounded and attains its bounds. (3) If $f$ is continuous on the interval $[a,b]$ and $f(a)<t$ and $f(b)>t,$ then there exists $c\in(a,b)$ such that $f(c)=t.$ (If there are obvious omissions from this list, please add them.)}}[/add]

[add]I want to find a continuous function $f$ that has certain properties.{{
*The advice here depends very much on the nature of the problem. At one extreme, the property required of the continuous function might trivially imply continuity, in which case the word "continuous" could just as well be omitted. So let's assume that the problem is one where $f$ is required to have a property that appears to be hard to reconcile with continuity. A first step would be to try out a few well-known continuous functions, but if the problem is an interesting one then this is almost guaranteed not to work. At that point, your main option is to build a continuous function in some way. An elementary way that sometimes works is to define your function separately on various intervals and make sure they match up at the endpoints. For example, piecewise linear functions are useful for some purposes. If that doesn't work, then a more advanced technique that is often used for this type is to build your function as a uniform limit of continuous functions with properties closer and closer to the property you are aiming for. See [[Constructing exotic sets and functions using limiting arguments]] for some examples of this technique. Sometimes this idea is combined with the previous one: for example, one might take a uniform limit of piecewise linear functions.}}[/add]

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