Continuity

Let \(f\) be a function. If \(f\) has \(f(x_0)\) has a limit at x_0, we say that \(f\) is continuous at \(x_0\). If \(A\) is a set of real numbers and \(f\) is continuous at every point of the set, we say that \(f\) is continuous on \(A\). This is especially appropriate when \(A\) is a segments: all points between \(a\) and \(b\).

Assume that \(f\) is continuous on \([a,b]\), all points between \(a\) and \(b\), inclusive. We want to show that there is some \(M\) such that \(|f(x)|M\), \(a\leq x_n\leq b\). Since \(x_n\) is a bounded sequence of real number, there is a convergent subsequence \(x_{n_k}\). We note that since \(n_k\geq k\), \(|f(x_{n_k})|>n_k\geq k\). \(x_{n_k}\) converges to \(x_0\), and so \(a\leq x_0\leq b\). Therefore, \(f(x_{n_k})\) converges to \(f(x_0)\). But \(f(x_{n_k})\) is not a bounded sequence, which is a contradiction. Therefore, \(f\) is bounded on \(A\).

Let us again assume that \(A\) is a segment, and \(f\) is continuous there. Assume that there is a \(\epsilon_0\) such that for every \(\delta\) there is an \(x, y\) such that \(|x-y|<\delta\) and \(|f(x)-f(y)|\geq \epsilon_0\). Now let us take \(x_n\) to be such an \(x\) for \(\delta=1/n\). \(x_n\) has a convergent subsequence, \(x_{n_k}\) converging to \(x_0\). We know that \(f\) is continuous at \(x_0\). Let us take a \(\delta\) corresponding to \(\epsilon_0/2\). We have that if \(|x-x_0|<\delta\) than \(|f(x)-f(x_0)|K\), \(|x_{n_k}-x_0`<\delta/2\). Now we know that for every \(x_{n_k}\) there is a \(y_k\) such that \(|x_{n_k}-y_k|K\) and such that \(1/k<\delta/2\), we have \(|y_k-x_0|<\delta\), so \(|f(y_k)-f(x_0)|<\epsilon/2\), and therefore we have \(|f(x_{n_k})-f(y_k)|=|f(x_{n_k})-f(x_0)+f(x_0)-f(y_k)|\leq |f(x_{n_k})-f(x_0)|+|f(x_0)-f(y_k)|<\epsilon_0/2+\epsilon_0/2=\epsilon_0\) in contradiction to the definition of \(k\). Therefore, there is no such \(\epsilon_0\), and so for every \(\epsilon\) there is a \(\delta\) such that if \(|x-y|<\delta\) then \(|f(x)-f(y)|<\epsilon\). In this case, we call \(f\) uniformly continuous on \(A\). As we have seen, every continuous function on a segment is uniformly continuous.

Assume \(f\) is a continuous function on a segment \(A\). We know that \(f\) is bounded from above. In other words, the set \(B={f(x):x\in A}\) has an upper bound, and therefore a least upper bound, say \(c\). In other words, if \(d<c\), there is an \(x\), \(d<f(x)\leq c\). Define \(x_n\) to be such an \(x\) for \(d=c-1/n\). Now we have \(x_{n_k}\) a converging subsequence, and \(c-1/k\leq c-1/n_k<f(x_{n_k})\leq c\). Assume that \(x_{n_k}\) converges to \(x_0\). Then \(f(x_{n_k})\) converges, by continuity, to f(x_0), and by sandwich, to \(c\). Therefore, \(f(x_0)=c\) and so the least upper bound is actually attained (in this case, we call it a maximum). For similar reasons, \(f\) has a minimum on \(A\).

Let \(f\) be a continuous function on a segment \(A=[a,b]\). If \(c\) is such that \(f(a)<c<f(b)\), then there is an \(x\), \(f(c)=x\). Let \(a_n, b_n\) be defined by \(a_0=a,b_0=b\). Let \(t=(a_n+b_n)/2\). If \(f(t)0\). \(t<t+\epsilon\), so \(x=f(t)0\), and for similar reasons \(\delta_2=f(t)-f(t-\epsilon)>0\). Define \(\delta\) to be the smaller of the two. If \(|x-y|<\delta\), then \(f(t-\epsilon)=f(t)-(f(t)-f(t-\epsilon))<x-\delta_2\leq x-\delta<y<x+\delta<x+\delta_1=f(t)+f(t+\epsilon)-f(t)=f(t+\epsilon)\). Applying \(f^{-1}\) to the inequality (since \(f^{-1}\) is increasing) gives us \(t-\epsilon<f^{-1}(y)<t+\epsilon\), or \(|f^{-1}(y)-f^{-1}(x)|0\). If \(a>0\) is a real number, then if we find \(M>a, M>1\), \(M^n>a\). \(0^n=00\), \(x^q\) is a well defined real function on the non-negative numbers (and therefore, if we define \(x^{-q}=1/x^q\)), we have \(x^q\) is a well defined function on the positive numbers for any rational \(q\)!