diff options
Diffstat (limited to 'šola')
| -rw-r--r-- | šola/ana1/teor.lyx | 1459 |
1 files changed, 1427 insertions, 32 deletions
diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx index 28f0397..2b6057d 100644 --- a/šola/ana1/teor.lyx +++ b/šola/ana1/teor.lyx @@ -2849,7 +2849,20 @@ Lahko se zgodi, \end_layout \begin_layout Theorem* -Naj bo +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hypertarget{kmoz}{Konvergenca monotonega in omejenega zaporedja} +\end_layout + +\end_inset + +. + Naj bo \begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$ \end_inset @@ -3862,7 +3875,20 @@ s čimer dobimo zgornjo mejo \end_inset , - je torej zaporedje omejeno in ker je tudi monotono po prejšnjem izreku konvergira. + je torej zaporedje omejeno in ker je tudi monotono po +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{kmoz}{prejšnjem izreku} +\end_layout + +\end_inset + + konvergira. \end_layout \end_deeper @@ -7157,7 +7183,7 @@ TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH, Ideja: Izdelati želimo formulacijo, s katero preverimo, - le lahko z dovolj majhno spremembo + če lahko z dovolj majhno spremembo \begin_inset Formula $x$ \end_inset @@ -7455,7 +7481,20 @@ okolice \end_deeper \begin_layout Theorem* -Naj bo +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hypertarget{kzsppom}{Karakterizacija zveznosti s pomočjo praslik odprtih množic} +\end_layout + +\end_inset + +. + Naj bo \begin_inset Formula $f:D\to\mathbb{R}$ \end_inset @@ -8111,6 +8150,329 @@ Naj bo \end_layout \begin_layout Example* +Kvadratna funkcija +\begin_inset Formula $f\left(x\right)=x^{2}$ +\end_inset + + je zvezna. + Vzemimo poljuben +\begin_inset Formula $a\in\mathbb{R},\varepsilon>0$ +\end_inset + +. + Obstajati mora taka +\begin_inset Formula $\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Podan imamo torej +\begin_inset Formula $a$ +\end_inset + + in +\begin_inset Formula $\varepsilon$ +\end_inset + +, + želimo najti +\begin_inset Formula $\delta$ +\end_inset + +. + Želimo priti do neenakosti, + ki ima na manjši strani +\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|x^{2}-a^{2}\right|$ +\end_inset + + in na večji strani nek izraz z +\begin_inset Formula $\left|x-a\right|$ +\end_inset + +, + da ta +\begin_inset Formula $\left|x-a\right|$ +\end_inset + + nadomestimo z +\begin_inset Formula $\delta$ +\end_inset + + in nato večjo stran enačimo z +\begin_inset Formula $\varepsilon$ +\end_inset + +, + da izrazimo +\begin_inset Formula $\varepsilon$ +\end_inset + + v odvisnosti od +\begin_inset Formula $\delta$ +\end_inset + + in +\begin_inset Formula $a$ +\end_inset + +. +\end_layout + +\begin_layout Example* +Računajmo: + +\begin_inset Formula $\left|x^{2}-a^{2}\right|=\left|x-a\right|\left|x+a\right|$ +\end_inset + +. + Predelajmo izraz +\begin_inset Formula $\left|x+a\right|=\left|\left(x-a\right)+2a\right|\leq\left|x-a\right|+\left|2a\right|$ +\end_inset + +, + torej skupaj +\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\left|x-a\right|\left(\left|x-a\right|+\left|2a\right|\right)$ +\end_inset + +. + Sedaj nadomestimo +\begin_inset Formula $\left|x-a\right|$ +\end_inset + + z +\begin_inset Formula $\delta$ +\end_inset + +: + +\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\delta\left(\delta+\left|2a\right|\right)$ +\end_inset + +. + Iščemo tak +\begin_inset Formula $\varepsilon$ +\end_inset + +, + da velja +\begin_inset Formula $\left|x^{2}-a^{2}\right|<\varepsilon$ +\end_inset + +, + zato enačimo +\begin_inset Formula $\delta\left(\delta+\left|2a\right|\right)=\varepsilon$ +\end_inset + + in dobimo kvadratno enačbo +\begin_inset Formula $\delta^{2}+\left|2a\right|\delta-\varepsilon=0$ +\end_inset + +, + ki jo rešimo z obrazcem za ničle: +\begin_inset Formula +\[ +\delta_{1,2}=\frac{-2\left|a\right|\pm\sqrt{4\left|a\right|^{2}-4\varepsilon}}{2}=-\left|a\right|\pm\sqrt{\left|a\right|^{2}-\varepsilon} +\] + +\end_inset + +Toda ker iščemo le pozitivne +\begin_inset Formula $\delta$ +\end_inset + +, + je edina rešitev +\begin_inset Formula +\[ +\delta=-\left|a\right|+\sqrt{\left|a\right|^{2}-\varepsilon}=\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|=\frac{\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|}{1}=\frac{\left(\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|\right)\left(\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|\right)}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}=\frac{\varepsilon}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|} +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Definition* +Naj bo +\begin_inset Formula $D\subset\mathbb{R},a\in\mathbb{R}\ni:\forall\varepsilon>0:D\cap\left(a,a+\varepsilon\right)\not=\emptyset$ +\end_inset + +. + Naj bo +\begin_inset Formula $f:D\to\mathbb{R}$ +\end_inset + +. + Število +\begin_inset Formula $L_{+}\in\mathbb{R}$ +\end_inset + + je desna limita funkcije +\begin_inset Formula $f$ +\end_inset + + v točki +\begin_inset Formula $a$ +\end_inset + +, + če +\begin_inset Formula $\forall\left(a_{n}\right)_{n\in\mathbb{N}}\subset D\cap\left(a,\infty\right):a_{n}\to a\Rightarrow f\left(a_{n}\right)\to L_{+}$ +\end_inset + + ZDB če za vsako k +\begin_inset Formula $a$ +\end_inset + + konvergentno zaporedje s členi desno od +\begin_inset Formula $a$ +\end_inset + + velja, + da funkcijske vrednosti členov konvergirajo k +\begin_inset Formula $L_{+}$ +\end_inset + +. + Oznaka +\begin_inset Formula $L_{+}=\lim_{x\to a^{+}}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)=f\left(a+0\right)$ +\end_inset + +. + Podobno definiramo tudi levo limito +\begin_inset Formula $L_{-}=\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=f\left(a-0\right)$ +\end_inset + +. +\end_layout + +\begin_layout Theorem* +Naj bo +\begin_inset Formula $D\subset\mathbb{R}$ +\end_inset + + in +\begin_inset Formula $a\in\mathbb{R}$ +\end_inset + + da velja +\begin_inset Formula $\forall\varepsilon>0:D\cap\left(a,a-\varepsilon\right)\not=\emptyset\wedge D\cap\left(a,a+\varepsilon\right)\not=\emptyset$ +\end_inset + +. + Naj bo +\begin_inset Formula $f:D\to\mathbb{R}$ +\end_inset + +. + Velja +\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$ +\end_inset + + V tem primeru velja +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Označimo +\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$ +\end_inset + +. + Če +\begin_inset Formula $\exists f\left(a+0\right)$ +\end_inset + + in +\begin_inset Formula $\exists f\left(a-0\right)$ +\end_inset + +, + vendar +\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$ +\end_inset + +, + pravimo, + da ima +\begin_inset Formula $f$ +\end_inset + + v točki +\begin_inset Formula $a$ +\end_inset + + +\begin_inset Quotes gld +\end_inset + +skok +\begin_inset Quotes grd +\end_inset + +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$ +\end_inset + + ne obstaja. + Zakaj? + Izračunajmo levo in desno limito: +\begin_inset Formula +\[ +\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1 +\] + +\end_inset + +Toda +\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$ +\end_inset + +. +\end_layout + +\begin_layout Definition* +Funkcija +\begin_inset Formula $f$ +\end_inset + + je na intervalu +\begin_inset Formula $D$ +\end_inset + + odsekoma zvezna, + če je zvezna povsod na +\begin_inset Formula $D$ +\end_inset + +, + razen morda v končno mnogo točkah, + v katerih ima skok. +\end_layout + +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + +\begin_layout Example* Naj bo \begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$ \end_inset @@ -8234,7 +8596,7 @@ Da naš sklep res potrdimo, \end_inset in -\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)=1$ +\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)\eqqcolon L$ \end_inset , @@ -8243,7 +8605,32 @@ Da naš sklep res potrdimo, \end_inset in -\begin_inset Formula $\lim_{x\to a}g\left(x\right)=1$ +\begin_inset Formula $\lim_{x\to a}g\left(x\right)=L$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Proof +Naj bo +\begin_inset Formula $A=A\left(x\right)\coloneqq\max\left\{ \left|f\left(x\right)-L\right|,\left|h\left(x\right)-L\right|\right\} $ +\end_inset + +. + Velja +\end_layout + +\begin_deeper +\begin_layout Itemize +\begin_inset Formula $g\left(x\right)-L\leq h\left(x\right)-L\leq\left|h\left(x\right)-L\right|\leq A\left(x\right)$ +\end_inset + + in +\end_layout + +\begin_layout Itemize +\begin_inset Formula $L-g\left(x\right)\leq L-f\left(x\right)\leq\left|f\left(x\right)-L\right|\leq A\left(x\right)$ \end_inset . @@ -8251,72 +8638,1080 @@ Da naš sklep res potrdimo, \end_deeper \begin_layout Proof -TODO XXX FIXME DOKAZ V SKRIPTi +Posledično +\begin_inset Formula $\left|g\left(x\right)-L\right|\leq A\left(x\right)$ +\end_inset + +. + Naj bo sedaj +\begin_inset Formula $\varepsilon>0$ +\end_inset + + poljuben. + Tedaj velja +\begin_inset Formula $\exists\delta_{1}>0\ni:\left|x-a\right|<\delta_{1}\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$ +\end_inset + + in +\begin_inset Formula $\exists\delta_{2}>0\ni:\left|x-a\right|<\delta_{2}\Rightarrow\left|h\left(x\right)-L\right|<\varepsilon$ +\end_inset + +. + Za +\begin_inset Formula $\delta\coloneqq\min\left\{ \delta_{1},\delta_{2}\right\} $ +\end_inset + + torej velja +\begin_inset Formula $\left|x-a\right|<\delta\Rightarrow\left|g\left(x\right)-L\right|<\varepsilon$ +\end_inset + +. +\end_layout + +\begin_layout Subsection +Zvezne funkcije na kompaktnih množicah +\end_layout + +\begin_layout Definition* +Množica +\begin_inset Formula $K\subseteq\mathbb{R}$ +\end_inset + + je kompaktna +\begin_inset Formula $\Leftrightarrow$ +\end_inset + + je zaprta in omejena ZDB je unija zaprtih intervalov. +\end_layout + +\begin_layout Theorem* +Naj bo +\begin_inset Formula $K\subset\mathbb{R}$ +\end_inset + + kompaktna in +\begin_inset Formula $f:K\to\mathbb{R}$ +\end_inset + + zvezna. + Tedaj je +\begin_inset Formula $f$ +\end_inset + + omejena in doseže minimum in maksimum. \end_layout \begin_layout Example* -\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$ +Primeri funkcij. +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $f_{1}\left(x\right)=\frac{1}{x}$ \end_inset - ne obstaja. - Zakaj? - Izračunajmo levo in desno limito: + na +\begin_inset Formula $I_{1}=(0,1]$ +\end_inset + +. + +\begin_inset Formula $f_{1}$ +\end_inset + + je zvezna in +\begin_inset Formula $\lim_{x\to0}f_{1}\left(x\right)=\infty$ +\end_inset + +, + torej ni omejena, + a +\begin_inset Formula $I_{1}$ +\end_inset + + ni zaprt. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $f_{2}\left(x\right)=\begin{cases} +0 & ;x=0\\ +\frac{1}{x} & ;x\in(0,1] +\end{cases}$ +\end_inset + + ni omejena in je definirana na kompaktni množici, + a ni zvezna. +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $f_{3}\left(x\right)=x$ +\end_inset + + na +\begin_inset Formula $x\in\left(0,1\right)$ +\end_inset + +. + Je omejena, + ne doseže maksimuma, + a +\begin_inset Formula $D_{f_{3}}$ +\end_inset + + ni kompaktna (ni zaprta). +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $f_{4}\left(x\right)=\begin{cases} +x & ;x\in\left(0,1\right)\\ +\frac{1}{2} & ;x\in\left\{ 0,1\right\} +\end{cases}$ +\end_inset + +. + Velja +\begin_inset Formula $\sup f_{4}=1$ +\end_inset + +, + ampak ga ne doseže, + a ni zvezna +\end_layout + +\end_deeper +\begin_layout Proof +Naj bo +\begin_inset Formula $K\subseteq\mathbb{R}$ +\end_inset + + kompaktna in +\begin_inset Formula $f:K\to\mathbb{R}$ +\end_inset + + zvezna. +\end_layout + +\begin_deeper +\begin_layout Itemize +Omejenost navzgor: + PDDRAA +\begin_inset Formula $f$ +\end_inset + + ni navzgor omejena. + Tedaj +\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\geq n$ +\end_inset + + (*). + Ker je +\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + omejeno zaporedje (vsi členi so na kompaktni +\begin_inset Formula $K$ +\end_inset + +), + ima stekališče, + recimo mu +\begin_inset Formula $s\in\mathbb{R}$ +\end_inset + +. + Vemo, + da tedaj obstaja podzaporedje +\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$ +\end_inset + +. + Ker je +\begin_inset Formula $K$ +\end_inset + + tudi zaprta, + sledi +\begin_inset Formula $s\in K$ +\end_inset + +. + Ker je +\begin_inset Formula $f$ +\end_inset + + zvezna na +\begin_inset Formula $K$ +\end_inset + +, + velja +\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$ +\end_inset + +. + Toda po (*) sledi +\begin_inset Formula $\lim_{k\to\infty}f\left(x_{n_{k}}\right)=\infty$ +\end_inset + +, + zato +\begin_inset Formula $f\left(s\right)=\infty$ +\end_inset + +, + kar ni mogoče, + saj je +\begin_inset Formula $f\left(s\right)\in\mathbb{R}$ +\end_inset + +. + +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +. + Torej je +\begin_inset Formula $f$ +\end_inset + + navzgor omejena. +\end_layout + +\begin_layout Itemize +Omejenost navzdol: + PDDRAA +\begin_inset Formula $f$ +\end_inset + + ni navzdol omejena. + Tedaj +\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\leq-n$ +\end_inset + + (*). + Ker je +\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + omejeno zaporedje (vsi členi so na kompaktni +\begin_inset Formula $K$ +\end_inset + +), + ima stekališče, + recimo mu +\begin_inset Formula $s\in\mathbb{R}$ +\end_inset + +. + Vemo, + da tedaj obstaja podzaporedje +\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$ +\end_inset + +. + Ker je +\begin_inset Formula $K$ +\end_inset + + tudi zaprta, + sledi +\begin_inset Formula $s\in K$ +\end_inset + +. + Ker je +\begin_inset Formula $f$ +\end_inset + + zvezna na +\begin_inset Formula $K$ +\end_inset + +, + velja +\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$ +\end_inset + +. + Toda po (*) sledi +\begin_inset Formula $\lim_{k\to\infty}f\left(s_{n_{k}}\right)=-\infty$ +\end_inset + +, + zato +\begin_inset Formula $f\left(s\right)=-\infty$ +\end_inset + +, + kar ni mogoče, + saj je +\begin_inset Formula $f\left(s\right)\in\mathbb{R}$ +\end_inset + +. + +\begin_inset Formula $\rightarrow\!\leftarrow$ +\end_inset + +. + Torej je +\begin_inset Formula $f$ +\end_inset + + navzgor omejena. + +\end_layout + +\begin_layout Itemize +Doseže maksimum: + Označimo +\begin_inset Formula $M\coloneqq\sup_{x\in K}f\left(x\right)$ +\end_inset + +. + Ravnokar smo dokazali, + da +\begin_inset Formula $M<\infty$ +\end_inset + +. + Po definiciji supremuma +\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)>M-\frac{1}{n}$ +\end_inset + +. + Ker je +\begin_inset Formula $K$ +\end_inset + + omejena, + ima +\begin_inset Formula $\left(t_{n}\right)_{n}$ +\end_inset + + stekališče in ker je zaprta, + velja +\begin_inset Formula $t\in K$ +\end_inset + +, + zato +\begin_inset Formula $\exists$ +\end_inset + + podzaporedje +\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$ +\end_inset + +. + Ker je +\begin_inset Formula $f$ +\end_inset + + zvezna, + velja +\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$ +\end_inset + +. + Toda ker +\begin_inset Formula $f\left(t_{n_{j}}\right)>M-\frac{1}{n_{j}}\geq M-\frac{1}{j}$ +\end_inset + +, + velja +\begin_inset Formula $f\left(t\right)\geq M$ +\end_inset + +. + Hkrati po definiciji +\begin_inset Formula $M$ +\end_inset + + velja +\begin_inset Formula $f\left(t\right)\leq M$ +\end_inset + +. + Sledi +\begin_inset Formula $M=f\left(t\right)$ +\end_inset + + in zato +\begin_inset Formula $M=\max_{x\in K}f\left(x\right)$ +\end_inset + +. +\end_layout + +\begin_layout Itemize +Doseže minimum: + Označimo +\begin_inset Formula $M\coloneqq\inf_{x\in K}f\left(x\right)$ +\end_inset + +. + Ko smo dokazali omejenost, + smo dokazali, + da +\begin_inset Formula $M>-\infty$ +\end_inset + +. + Po definiciji infimuma +\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)<M+\frac{1}{n}$ +\end_inset + +. + Ker je +\begin_inset Formula $K$ +\end_inset + + omejena, + ima +\begin_inset Formula $\left(t_{n}\right)_{n}$ +\end_inset + + stekališče in ker je zaprta, + velja +\begin_inset Formula $t\in K$ +\end_inset + +, + zato +\begin_inset Formula $\exists$ +\end_inset + + podzaporedje +\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$ +\end_inset + +. + Ker je +\begin_inset Formula $f$ +\end_inset + + zvezna, + velja +\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$ +\end_inset + +. + Toda ker +\begin_inset Formula $f\left(t_{n_{j}}\right)<M-\frac{1}{n_{j}}\leq M-\frac{1}{j}$ +\end_inset + +, + velja +\begin_inset Formula $f\left(t\right)\leq M$ +\end_inset + +. + Hkrati po definiciji +\begin_inset Formula $M$ +\end_inset + + velja +\begin_inset Formula $f\left(t\right)\geq M$ +\end_inset + +. + Sledi +\begin_inset Formula $M=f\left(t\right)$ +\end_inset + + in zato +\begin_inset Formula $M=\min_{x\in K}f\left(x\right)$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Theorem* +Naj bo +\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$ +\end_inset + + zvezna in +\begin_inset Formula $f\left(a\right)f\left(b\right)<0$ +\end_inset + +. + Tedaj +\begin_inset Formula $\exists\xi\in\left(a,b\right)\ni:f\left(\xi\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Interval +\begin_inset Formula $I_{0}=\left[a,b\right]$ +\end_inset + + razpolovimo. + To pomeni, + da pogledamo levo in desno polovico intervala +\begin_inset Formula $I_{0}$ +\end_inset + +, + torej +\begin_inset Formula $\left[a,\frac{a+b}{2}\right]$ +\end_inset + + in +\begin_inset Formula $\left[\frac{a+b}{2},b\right]$ +\end_inset + +. + Če je +\begin_inset Formula $f\left(\frac{a+b}{2}\right)=0$ +\end_inset + +, + smo našli iskano točko +\begin_inset Formula $\xi$ +\end_inset + +, + sicer z +\begin_inset Formula $I_{1}$ +\end_inset + + označimo katerokoli izmed polovic, + ki ima +\begin_inset Formula $f$ +\end_inset + + v krajiščih različno predznačene funkcijske vrednosti. + Torej +\begin_inset Formula $I_{1}=\begin{cases} +\left[a,\frac{a+b}{2}\right] & ;f\left(a\right)f\left(\frac{a+b}{2}\right)<0\\ +\left[\frac{a+b}{2},b\right] & ;f\left(\frac{a+b}{2}\right)f\left(b\right)<0 +\end{cases}$ +\end_inset + +. + S postopkom nadaljujemo. + Če v končno mnogo korakih najdemo +\begin_inset Formula $\xi$ +\end_inset + +, + da je +\begin_inset Formula $f\left(\xi\right)=0$ +\end_inset + +, + fino, + sicer pa dobimo zaporedje intervalov +\begin_inset Formula $I_{n}=\left[a_{n},b_{n}\right]\subset\left[a,b\right]=I_{0}\ni:$ +\end_inset + + +\end_layout + +\begin_deeper +\begin_layout Enumerate +\begin_inset Formula $\forall n\in\mathbb{N}:\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$ +\end_inset + + in +\end_layout + +\begin_layout Enumerate +\begin_inset Formula $\forall n\in\mathbb{N}:I_{n+1}\subset I_{n}$ +\end_inset + +, + torej +\begin_inset Formula $a_{n+1}\geq a_{n}\wedge b_{n+1}\leq b_{n}$ +\end_inset + +, + in +\end_layout + +\begin_layout Enumerate +\begin_inset CommandInset label +LatexCommand label +name "enu:različni-predznaki-istoležnih-clenov" + +\end_inset + + +\begin_inset Formula $\forall n\in\mathbb{N}:f\left(a_{n}\right)f\left(b_{n}\right)<0$ +\end_inset + +. +\end_layout + +\end_deeper +\begin_layout Proof +Ker sta zaporedji +\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + in +\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$ +\end_inset + + omejeni in monotoni, + imata po +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{kmoz}{izreku o konvergenci monotonih in omejenih zaporedij} +\end_layout + +\end_inset + + limiti +\begin_inset Formula $\alpha\coloneqq\lim_{n\to\infty}a_{n}=\sup_{n\in\mathbb{N}}a_{n}$ +\end_inset + + in +\begin_inset Formula $\beta\coloneqq\lim_{n\to\infty}=\sup_{n\in\mathbb{N}}b_{n}$ +\end_inset + + in +\begin_inset Formula $\alpha,\beta\in I_{0}$ +\end_inset + +, + ker je +\begin_inset Formula $I_{0}$ +\end_inset + + zaprt. +\end_layout + +\begin_layout Proof +Sledi +\begin_inset Formula $\forall n\in\mathbb{N}:\left|\alpha-\beta\right|=\beta-\alpha\leq b_{n}-a_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$ +\end_inset + +, + torej +\begin_inset Formula $\lim_{n\to\infty}\left|\alpha-\beta\right|=0\Rightarrow\alpha-\beta=0$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Ker je +\begin_inset Formula $f$ +\end_inset + + zvezna in +\begin_inset Formula $a_{n},b_{n},\xi\in I_{0}$ +\end_inset + +, + sledi +\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(\alpha\right)=f\left(\xi\right)=f\left(\beta\right)=\lim_{n\to\infty}f\left(b_{n}\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Po točki +\begin_inset CommandInset ref +LatexCommand ref +reference "enu:različni-predznaki-istoležnih-clenov" +plural "false" +caps "false" +noprefix "false" +nolink "false" + +\end_inset + + velja +\begin_inset Formula $f\left(\alpha\right)f\left(\beta\right)\leq0$ +\end_inset + +. + Ker pa +\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)$ +\end_inset + +, + velja +\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)=f\left(\xi\right)=0$ +\end_inset + +. +\end_layout + +\begin_layout Corollary* +Naj bo +\begin_inset Formula $I=\left[a,b\right]$ +\end_inset + + omejen zaprt interval +\begin_inset Formula $\in\mathbb{R}$ +\end_inset + + in +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + zvezna. + Tedaj +\begin_inset Formula $\exists x_{-},x_{+}\in I\ni:\forall x\in I:f\left(x\right)\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$ +\end_inset + + in +\begin_inset Formula $\forall y\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]\exists x\in I\ni:y=f\left(x\right)$ +\end_inset + + ZDB +\begin_inset Formula $f\left(I\right)=\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$ +\end_inset + + ZDB zvezna funkcija na zaprtem intervalu +\begin_inset Formula $\left[a,b\right]$ +\end_inset + + doseže vse funkcijske vrednosti na intervalu +\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Dokaz posledice. + Naj bo +\begin_inset Formula $y$ +\end_inset + + poljuben. + Če je +\begin_inset Formula $y=f\left(x_{-}\right)$ +\end_inset + +, + smo našli +\begin_inset Formula $x=x_{-}$ +\end_inset + +. + Če je +\begin_inset Formula $y=f\left(x_{+}\right)$ +\end_inset + +, + smo našli +\begin_inset Formula $x=x_{+}$ +\end_inset + +. + Sicer pa je +\begin_inset Formula $f\left(x_{-}\right)<y<f\left(x_{+}\right)$ +\end_inset + +. + Oglejmo si funkcijo +\begin_inset Formula $g\left(x\right)\coloneqq f\left(x\right)-y$ +\end_inset + +. + Ker je +\begin_inset Formula $g\left(x_{-}\right)=f\left(x_{-}\right)-y<0$ +\end_inset + + in +\begin_inset Formula $g\left(x_{+}\right)=f\left(x_{+}\right)-y>0$ +\end_inset + + in +\begin_inset Formula $g$ +\end_inset + + zvezna na +\begin_inset Formula $\left[x_{-}-y,x_{+}-y\right]$ +\end_inset + +, + torej po prejšnjem izreku +\begin_inset Formula $\exists x\in\left[x_{-}-y,x_{+}-y\right]\ni:g\left(x\right)=0$ +\end_inset + +, + kar pomeni ravno +\begin_inset Formula $f\left(x\right)=y$ +\end_inset + +. +\end_layout + +\begin_layout Theorem* +Naj bo +\begin_inset Formula $I$ +\end_inset + + poljuben interval med +\begin_inset Formula $a,b\in\mathbb{R}\cup\left\{ -\infty,\infty\right\} $ +\end_inset + + in +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + zvezna in strogo monotona. + Tedaj je +\begin_inset Formula $f\left(I\right)$ +\end_inset + + interval med +\begin_inset Formula $f\left(a+0\right)$ +\end_inset + + in +\begin_inset Formula $f\left(a-0\right)$ +\end_inset + +. + Inverzna funkcija +\begin_inset Formula $f^{-1}$ +\end_inset + + je definirana na +\begin_inset Formula $f\left(I\right)$ +\end_inset + + in zvezna. +\end_layout + +\begin_layout Example* +\begin_inset Formula $f\coloneqq\arctan$ +\end_inset + +, + +\begin_inset Formula $I\coloneqq\left(-\infty,\infty\right)$ +\end_inset + +, + zvezna. + Naj bo +\begin_inset Formula $y\in f\left(I\right)$ +\end_inset + + poljuben. + Tedaj +\begin_inset Formula $\exists!x\in I\ni:y=f\left(x\right)$ +\end_inset + + in definiramo +\begin_inset Formula $x\coloneqq f^{-1}\left(x\right)$ +\end_inset + +. + +\begin_inset Formula $f^{-1}$ +\end_inset + + obstaja in je spet zvezna. +\end_layout + +\begin_layout Proof +Ne bomo dokazali. +\begin_inset Note Note +status open + +\begin_layout Plain Layout +Označimo +\begin_inset Formula $g=f^{-1}:f\left(I\right)\to\mathbb{R}$ +\end_inset + +. + Uporabimo +\begin_inset ERT +status open + +\begin_layout Plain Layout + + +\backslash +hyperlink{kzsppom}{karakterizacijo zveznosti s pomočjo praslik odprtih množic} +\end_layout + +\end_inset + +. + Dokazujemo torej, + da +\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:g^{-1}\left(V\right)$ +\end_inset + + je zopet odprta množica +\begin_inset Formula $\subseteq f\left(I\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Velja +\begin_inset Formula $g^{-1}\left(V\right)=\left\{ x\in f\left(I\right);g\left(x\right)\in V\right\} =\left\{ x\in f\left(I\right):\exists v\in V\cap I\ni:x=f\left(v\right)\right\} =f\left(V\cap I\right)$ +\end_inset + +. +\end_layout + +\begin_layout Proof +Torej dokazujemo +\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:f\left(I\cap V\right)$ +\end_inset + + je spet zopet odprta +\begin_inset Formula $\subseteq f\left(I\right)$ +\end_inset + +, + kar je ekvivalentno \begin_inset Formula \[ -\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1 +\forall y\in f\left(I\cap V\right)\exists\delta>0\ni:\left(y-\delta,y+\delta\right)\cap f\left(I\right)\subset f\left(I\cap V\right). \] \end_inset -Toda -\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$ +Pišimo +\begin_inset Formula $y=f\left(x\right),x\in I\cap V$ \end_inset . + Privzemimo, + da +\begin_inset Formula $f$ +\end_inset + + narašča (če pada, + ravnamo podobno). + Ker jer +\begin_inset Formula $ $ +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + +\begin_layout Subsection +Enakomerna zveznost \end_layout \begin_layout Definition* -Označimo -\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$ +\begin_inset Formula $f:I\to\mathbb{R}$ \end_inset -. - Če -\begin_inset Formula $\exists f\left(a+0\right)$ + je enakomerno zvezna na +\begin_inset Formula $I$ \end_inset - in -\begin_inset Formula $\exists f\left(a-0\right)$ +, + če +\begin_inset Formula +\[ +\forall\varepsilon>0\exists\delta>0\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon. +\] + +\end_inset + + +\end_layout + +\begin_layout Note* +Primerjajmo to z definicijo +\begin_inset Formula $f:I\to\mathbb{R}$ +\end_inset + + je (nenujno enakomerno) zvezna na +\begin_inset Formula $I$ \end_inset , - vendar -\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$ + če +\begin_inset Formula +\[ +\forall\varepsilon>0,a\in I\exists\delta>0\forall x\in I:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon. +\] + +\end_inset + +Pri slednji definiciji je +\begin_inset Formula $\delta$ +\end_inset + + odvisna od +\begin_inset Formula $\varepsilon$ +\end_inset + + in +\begin_inset Formula $a$ \end_inset , - pravimo, - da ima -\begin_inset Formula $f$ + pri enakomerni zveznosti pa le od +\begin_inset Formula $\varepsilon$ \end_inset - v točki +. +\end_layout + +\begin_layout Example* +\begin_inset Formula $f\left(x\right)=\frac{1}{x}$ +\end_inset + + ni enakomerno zvezna, + ker je +\begin_inset Formula $\delta$ +\end_inset + + odvisen od \begin_inset Formula $a$ \end_inset - -\begin_inset Quotes gld +. + Če pri fiksnem +\begin_inset Formula $\varepsilon$ \end_inset -skok -\begin_inset Quotes grd + pomaknemo tisto pozitivno točko, + v kateri preizkušamo zveznost, + bolj v levo, + bo na neki točki potreben ožji, + manjši +\begin_inset Formula $\delta$ \end_inset . \end_layout +\begin_layout Standard +\begin_inset Separator plain +\end_inset + + +\end_layout + \begin_layout Corollary* -sssssssssss +sssssssssss \end_layout \begin_layout Corollary* |