Continuity

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Functions of one variable

Definition

Definition : Continuity at one point

Let $f$ be a function defined on an interval $I$ of $\mathbb R$ and has values in $\mathbb R$. Let $x_0$ be an element of $I$. The function $f$ is said to be continuous in $x_0$ if :

$$\forall \epsilon > 0, \exists \delta > 0, \forall x \in I, |x_0 - x| < \delta \Longrightarrow |f(x_0) - f(x)| < \epsilon$$

Definition : Continuity

Let $f$ be a function defined on an interval $I$ of $\mathbb R$ and has values in $\mathbb R$. The function $f$ is said to be continuous if for all $x$ in $\mathbb R$, $f$ is continuous in $x$.

Results

Proposition

Let $f,g : I \subset \mathbb R \rightarrow \mathbb R$ be two continuous functions at a point $x_0 \in I$. Then :

• $\lambda \cdot f$ is continuous at $x_0$ (for all $\lambda \in \mathbb R$),
• $f + g$ is continuous at $x_0$,
• $f \times g$ is continuous at $x_0$,
• if $f(x_0) \neq 0$, then frac{1}{f} is continuous at $x_0$.

Theorem : Intermediate value theorem

Let $f$ be a continuous function on $\mathbb R$. Let $I$ be an interval of $\mathbb R$ and let $a, b \in I$ such that $a < b$ and $f(a)f(b) < 0$. Then the equation $f(x) = 0$ has at least one solution on the interval $[a, b] \in I$.

Proof

Even if we consider the function $−f$, we can assume without loss of generality that $f(a) < 0 < f(b)$ (the case where $f(a)=0$ or $f(b)=0$ being direct). Let $\mathcal A={x \ in [a,b] | f(x) < 0}$. Then $\mathcal A$ is not empty and upper bounded (because $a \in \mathcal A$ and is upper bounded by $b$). Thus, the set $\mathcal A$ has an upper bound. We note it $\alpha$. Because of the continuity of $f$ on the interval $I$, there exists $\eta > 0$ such that :

$$\forall x \in [a, a+\eta] \subset [a, b], f(x) \leq 0$$

and

$$\forall x \in [b - \eta, b] \subset [a, b], f(x) > 0$$

We then have $a < a + \eta \leq \alpha \leq b - \eta < b$. Thus, we have $\alpha \in ]a, b[$ and therefore there exists an integer $n_0$ such that :

$$\left ]\alpha - \frac{1}{n_0}, \alpha + \frac{1}{n_0} \right [ \subset ]a, b[$$

We therefore have for all $n > n_0$ that there exist real numbers $x_n \in \left ]\alpha, \alpha - \frac{1}{n_0} \right [$ and $y_n \in \left ]\alpha, \alpha + \frac{1}{n_0} \right [$. We have built two sequences which converge towards $\alpha$. Furthermore, we have by construction that :

$$f(\alpha) = lim_{n \rightarrow + \infty} f(x_n) \leq 0 \leq lim_{n \rightarrow + \infty} f(y_n) = f(\alpha)$$

We then deduce that $f(\alpha) = 0$. This completes the proof.

Source of proof : progresser-en-maths.com

Theorem : Bounds reached

If $f : [a, b] \rightarrow \mathbb R$ is a continuous function, then $f$ is bounded and reaches its bounds.

Proof

We suppose that $f$ is not bounded. Then $\forall A > 0, \exists x \in [a,b], f(x) > A$.

By taking $A = n$, we get $\exists x_n \in [a,b], f(x_n) > n$.

Therefore $lim_{n \rightarrow + \infty} f(x_n) = + \infty$.

According to the theorem of Bolzano-Weiestrass, there exists $\phi : \mathbb N \rightarrow \mathbb N$ strictly increasing such that $x_{\phi(n)}$ converges towards a limit $l$.

Then, by continuity, $lim_{n \rightarrow + \infty} f(x_{\phi(n)}) = f(l) \neq + \infty$ We therefore end up with a contradiction. Thus, $f$ is increased.

The set $f([a,b])$ being upper bounded it admits an upper bound which we will note $M$. We therefore have:

$$\forall \epsilon > 0, \exists x \in [a,b], M-\epsilon < f(x) \leq M$$

Let's take $\epsilon = \frac{1}{n}$. We therefore have, $\exists x_n \in [a,b], M - \frac{1}{n} < f(x_n) \leq M$.

We then obtain, according to the squeeze theorem, that, $lim_{n \rightarrow + \infty} f(x_n) = M$.

Furthermore, according to the theorem of Bolzano-Weierstrass, we have the existence of $\phi : \mathbb N \rightarrow \mathbb N$ strictly increasing such that $x_{\phi(n)}$ converges towards a limit $l$.

And since $f$ is continuous, we have $lim_{n \rightarrow + \infty} f(x_n) = f(l)$. Hence, $f(l) = M$. $f$ reaches therefore its upper bound $l$ in $M$.

Similarly, we can do the same thing on the lower bound.

Source of proof : progresser-en-maths.com