How can I prove that this function is continuous on the domain $[a,b]$?

Question

$f$ is a function defined on the domain $[a,b]$ and the following is true for each real number $x\in[a,b]$ then :

\begin{equation} \text{where: $\,\alpha\,$ is a real number of $[a,b]\,$}\begin{cases} f(x)\in[a,b] \\ \vert f(x)-f(\alpha)\vert \leqslant\vert x-\alpha\vert \end{cases} \end{equation}

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How do I prove that the function $f$ is continuous at $\,\alpha\,$ and then continuous on the domain $[a,b]\;$?

And how do I prove that the equation $\,f(x)=x\,$ accepts at least a solution in the domain $[a;b]\;$?

I know it's easy but i didn't understand what i have to do. How i'm supposed to prove it without the function expression.

I mean what i know about continuity is that: "Saying that the function f is continuous at $\,\alpha\,$ means that the limit of the function at $\,\alpha\,$ is f($\,\alpha\,$)"

(And the surprise i've been asked this question in the exam. And for clarification this is for high school so please simplify it, or don't use methods i don't know or didn't study)

Answer 1

EDIT: here is a proof using the following continuity definition:
$f$ is continuous on $a$ iff $\lim_{x \to a} (f(x))=f(a)$.

When $x \to \alpha, |x-\alpha| \to 0$, so
$|f(x)-f(\alpha)| \le |x-a| \to 0$, so
$f(x) \to f(a)$

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Assuming the continuity definition you know is the one with $\varepsilon$ and $\delta$, i.e. $f$ is continuous on $x_0$ iff
$\forall \varepsilon > 0, \exists \delta > 0, \forall x, |x-x_0| < \delta \Rightarrow |f(x)-f(x_0)| < \varepsilon$

$f$ is continuous at $\alpha$: apply the continuity definition with $x_0 = \alpha$ and $\delta = \varepsilon$. We have
$\forall \varepsilon > 0, \exists \delta > 0, \delta=\varepsilon, \forall x, |x-\alpha| < \delta \Rightarrow |f(x)-f(\alpha)| \le |x-\alpha| < \delta = \varepsilon$

$f$ is continuous on $[a,b]$:

• If the property with $\alpha$ is true $\forall \alpha \in [a,b]$, then $f$ is continuous on each point of $[a,b]$, which by definition means $f$ is continuous on $[a,b]$.
• If however $\alpha$ is one specific value in $[a,b]$, then that's wrong. Counterexample: $f$ defined on $[-1,1], \forall x \in [-1,0.5], f(x)=0,$ and $\forall x \in \; ]0.5,1], f(x)=0.5$. Then the property is true with $\alpha=0$, but $f$ is discontinuous on $0.5$.

The equation $f(x)=x$ has at least one solution on $[a,b]$: if $f$ is continuous on $[a,b]$, pose $g(x)=f(x)-x$, then $g$ is continuous on $[a,b]$ (sum of continuous functions), $g(a) = f(a)-a \ge 0$ and $g(b)=f(b)-b \le 0$, so by the intermediate value theorem, $\exists x_0 \in [a,b], g(x_0)=0$, so $f(x_0)=x_0$.

Note: the property $\forall x, |f(x)-f(\alpha)| \le K |x-\alpha|$ is called "$f$ is $K$-Lipschitz on $\alpha$". This is stronger than being continuous on $\alpha$. Here our function is $1$-Lipschitz on $\alpha$ (and so is $1$-Lipschitz on $[a,b]$ if the property is true $\forall \alpha \in [a,b]$).