# Real Analysis/Continuity

 Real Analysis Continuity

Now that we've defined the limit of a function, we're in a position to define what it means for a function to be continuous. The notion of Continuity captures the intuitive picture of a function "having no sudden jumps or oscillations". We will see several examples of discontinuous functions that illustrate the meaning of the definition. The idea of continuous functions is found in several areas of mathematics, apart from real analysis.

## Definition

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$
Let $c\in A$

We say that $f(x)$ is continuous at $c$ if and only if
For every $\varepsilon>0$, there exists $\delta>0$ such that $|x-c|<\delta\implies |f(x)-f(c)|<\varepsilon$.
We say $f$ itself is continuous if this condition holds for all points in $A$.

If $A$ is a union of intervals, the statement is equivalent to saying that $\lim_{x\to c}f(x)=f(c)$.

## Algebraic Operations

Since limits are preserved under algebraic operations, we see that if $f(x)$ and $g(x)$ are both continuous at c:

• $af(x)$ is continuous at $c$ for all $a \in \mathbb{R}$.
• $f(x) + g(x)$ is continuous at $c$.
• $f(x)g(x)$ is continuous at $c$.
• $f(x)/g(x)$ is continuous at $c$, assuming $g(c)$ is non-zero.

We can use sequential limits to prove that functions are discontinuous as follows:

• $f(x)$ is discontinuous at $c$ if and only if there are two sequences $(x_n)\rightarrow c$ and $(y_n)\rightarrow c$ such that $\lim_{n \rightarrow \infty}(f(x_n)) \not= \lim_{n \rightarrow \infty}(f(y_n))$.

## Composition

Another result that will allow us to construct many examples of continuous functions is that any composition of continuous functions is itself continuous:

### Theorem

If $f:B\rightarrow \mathbb{R}$ and $g:A\rightarrow B$ are continuous, then the composition $(f \circ g)(x) = f(g(x))$ is continuous on A.

#### Proof

Let $\epsilon>0$. Let $c \in A$.

Since f is continuous, $\exists \delta_1 > 0: |x-c|<\delta_1 \implies |f(x)-f(c)|<\epsilon$.

Since g is continuous, $\exists \delta_2 > 0: |x-c|<\delta_2 \implies |g(x)-g(c)|<\delta_1$.

Thus $|x-c|<\delta_2 \implies |g(x)-g(c)|<\delta_1 \implies |f(g(x))-f(g(c))|<\epsilon$, so $(f \circ g)(x)$ is continuous on A.

## The Intermediate Value Theorem

This is the big theorem on continuity. Essentially it says that continuous functions have no sudden jumps or breaks.

### Theorem (Intermediate Value Theorem)

Let f(x) be a continuous function. If $a and $f(a), then $\exists c \in (a,b): f(c) = m$.

#### Proof

Let $S = \{x \in (a,b): f(x) < m\}$, and let $c = \sup S$.

Let $\epsilon = |f(c) - m|$. By continuity, $\exists \delta: |x-c|< \delta \implies |f(x)-f(c)|< \epsilon$.

If f(c) < m, then $|f(c+\frac{\delta}{2}) - f(c)| < \epsilon$, so $f(c+\frac{\delta}{2}) < f(c) + \epsilon = m$. But then $c + \frac{\delta}{2} \in S$, which implies that c is not an upper bound for S, a contradiction.

If f(c) > m, then since $c = \sup S$, $\exists x: x \in S, c>x>c-\delta$. But since $|x-c|<\delta$, $|f(x)-f(c)|<\epsilon$, so $f(x)> f(c) - \epsilon$ = m, which implies that $x \notin S$, a contradiction. $\Box$

We will now prove the Minimum-Maximum theorem, which is another significant result that is related to continuity. Essentially, it states that any continuous image of a closed interval is bounded, and also that it attains these bounds.

## Minimum - Maximum Theorem

Let $f:[a,b]\to\mathbb{R}$ be continuous

Then
(i)$f([a,b])$ is bounded

(ii)If $M,m$ are respectively the upper and lower bounds of $f([a,b])$, then there exist $c,d\in [a,b]$ such that $f(c)=M,f(d)=m$

### Proof

(i)Assume if possible that $f$ is unbounded.

Let $x_1=\tfrac{a+b}{2}$. Then, $f$ is unbounded on at least one of the closed intervals $[a,x_1]$ and $[x_1,b]$ (for otherwise, $f$ would be bounded on $[a,b]$ contradicting the assumption). Call this interval $I_1$.

Similarly, partition $I_1$ into two closed intervals and let $I_2$ be the one on which $f$ is unbounded.

Thus we have a sequence of nested closed intervals $[a,b]\supseteq I_1\supseteq I_2\supseteq\ldots$ such that $f$ is unbounded on each of them.

We know that the intersection of a sequence of nested closed intervals is nonempty. Hence, let $x_0\in I_1\cap I_2\cap\ldots$

As $f(x)$ is continuous at $x=x_0$, there exists $\delta >0$ such that $x\in V_{\delta}(x_0)\implies f(x)\in (f(x_0)-1,f(x_0)+1)$ But by definition, there always exists $k\in\mathbb{N}$ such that $I_k\subseteq V_{\delta}(x_0)$, contradicting the assumption that $f$ is unbounded over $I_k$. Thus, $f$ is bounded over $[a,b]$

(ii) Assume if possible, $M=\sup (f([a,b]))$ but $M\notin f([a,b])$.

Consider the function $g(x)=\frac{1}{M-f(x)}$. By algebraic properties of continuity, $g:[a,b]\to\mathbb{R}$ is continuous. However, $M$ being a cluster point of $f([a,b])$, $g(x)$ is unbounded over $[a,b$, contradicting (i). Hence, $M\in f([a,b])$. Similarly, we can show that $m\in f([a,b])$.

## General Use

As mentioned, the idea of continuous functions is used in several areas of mathematics, most notably in Topology. A different characterization of continuity is useful in such scenarios.

### Theorem

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$

$f(x)$ is continuous at $x=c$ if and only if for every open neighbourhood $V$ of $f(x)$, there exists an open neighbourhood $U$ of $x$ such that $U\subseteq f^{-1}(V)$

It must be mentioned here that the term "Open Set" can be defined in much more general settings than the set of reals or even metric spaces, and hence the utility of this characterization.

## Uniform Continuity

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$

We say that $f$ is Uniformly Continuous on $A$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that if $x,y\in A$ and $|x-y|<\delta$ then $|f(x)-f(y)|<\varepsilon$

## Lipschitz continuity

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$

We say that $f$ is Lipschitz continuous on $A$ if and only if there exists a positive real constant $K$ such that, for all $x,y\in A$, $|f(x)-f(y)| \le K |x-y|$.

The smallest such $K$ is called the Lipschitz constant of the function $f$.