Continuity Theorems

Extreme Value Theorem

If $f$ is continuous on $[a,b]$, $f$ has a maximum and a minimum in $[a,b]$.

Proof Hint

Proof is quite hard.

Intermediate Value Theorem

Let $f$ is continuous on $[a,b]$. If $\exists u$ such that $f(a)>u>f(b)$ or $f(a)<u<f(b)$: $\exists c \in (a,b)$ such that $f(c)=u$.

Proof Hint

• Define $g(x)=f(x)-u$
• Start with $f(b)\lt u \lt f(a)$
• Define $A=\{ x \in [a,b] \,|\,g(x)\gt 0 \}$
• Show that $\sup A$ ($=c$) exists.
• Use the continuity definition. Assume and contradict these cases:
  • $c=a$ (use $2\epsilon = g(a)$)
  • $c=b$ (use $2\epsilon = -g(b)$)
  • $c\in(a,b)$ then contradict:
    • $g(c) \gt 0$ (similar to $c=a$ case)
    • $g(c) \lt 0$ (similar to $c=b$ case)

Sandwich (or Squeeze) Theorem

Let:

• For some $\delta>0$: $\forall x (0<|x-a|<\delta \implies f(x)\le g(x) \le h(x) )$
• $\lim_{x\to a} f(x) = \lim_{x\to a} h(x) = L \in \mathbb{R}$

Then $\lim_{x\to a} g(x) = L$. $$

Works for any kind of x limits.

Proof Hint

• Consider the limit definition of both $f$ and $h$ at $x=a$
• Show that $f$ is bounded below by $L-\epsilon$
• Show that $h$ is bounded above by $L+\epsilon$
• That concludes $g(x)-L$ is bounded by $\epsilon$ which leads to the required limit

”No sudden changes”

Positive

Let $f$ be continuous on $a$ and $f(a)>0$

$\implies \exists \delta >0; \forall x\,(|x-a|\lt\delta \implies f(x)>0)$ $$

Proof Hint

Take $\epsilon=\frac{f(a)}{2}$ $$

Negative

Let $f$ be continuous on $a$ and $f(a)<0$

$\implies \exists \delta >0; \forall x\,(|x-a|\lt\delta \implies f(x)<0)$ $$

Proof Hint

Take $\epsilon=-\frac{f(a)}{2}$ $$