Uniform convergence and inequalities











up vote
0
down vote

favorite












Assume that $f:,mathbb{R}rightarrowmathbb{R}$ is a continuous function. Assume also that exists a $M>0$ such that, for all $x in mathbb{R}$, we have $$left|fleft(xright)right|leq M.$$ Suppose that exists a sequence of function $f_n$ such that $f_n rightarrow f$ uniformly. I want to prove that, for all $x in mathbb{R}$, we have $$left|f_{n}left(xright)right|leq M$$ for every sufficiently large $n$.



My attempt. Since $f_n rightarrow f$ uniformly for all $epsilon>0$ exists a $N>0$ such that for all $x in mathbb{R}$ and $n geq N$ we have $$left|f_{n}left(xright)-fleft(xright)right|<epsilon.$$ So $$left|f_{n}left(xright)right|<left|fleft(xright)right|+left|f_{n}left(xright)-fleft(xright)right|$$ $$leq M + epsilon.$$ Is this proof correct? Thank you.






calculus real-analysis proof-verification inequality uniform-convergence







share|cite|improve this question


















  • 1




    What if $ f equiv M$ and $f_n equiv M+1/n$ ?
    – nicomezi
    Nov 22 at 11:54















up vote
0
down vote

favorite












Assume that $f:,mathbb{R}rightarrowmathbb{R}$ is a continuous function. Assume also that exists a $M>0$ such that, for all $x in mathbb{R}$, we have $$left|fleft(xright)right|leq M.$$ Suppose that exists a sequence of function $f_n$ such that $f_n rightarrow f$ uniformly. I want to prove that, for all $x in mathbb{R}$, we have $$left|f_{n}left(xright)right|leq M$$ for every sufficiently large $n$.



My attempt. Since $f_n rightarrow f$ uniformly for all $epsilon>0$ exists a $N>0$ such that for all $x in mathbb{R}$ and $n geq N$ we have $$left|f_{n}left(xright)-fleft(xright)right|<epsilon.$$ So $$left|f_{n}left(xright)right|<left|fleft(xright)right|+left|f_{n}left(xright)-fleft(xright)right|$$ $$leq M + epsilon.$$ Is this proof correct? Thank you.






calculus real-analysis proof-verification inequality uniform-convergence







share|cite|improve this question


















  • 1




    What if $ f equiv M$ and $f_n equiv M+1/n$ ?
    – nicomezi
    Nov 22 at 11:54













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Assume that $f:,mathbb{R}rightarrowmathbb{R}$ is a continuous function. Assume also that exists a $M>0$ such that, for all $x in mathbb{R}$, we have $$left|fleft(xright)right|leq M.$$ Suppose that exists a sequence of function $f_n$ such that $f_n rightarrow f$ uniformly. I want to prove that, for all $x in mathbb{R}$, we have $$left|f_{n}left(xright)right|leq M$$ for every sufficiently large $n$.



My attempt. Since $f_n rightarrow f$ uniformly for all $epsilon>0$ exists a $N>0$ such that for all $x in mathbb{R}$ and $n geq N$ we have $$left|f_{n}left(xright)-fleft(xright)right|<epsilon.$$ So $$left|f_{n}left(xright)right|<left|fleft(xright)right|+left|f_{n}left(xright)-fleft(xright)right|$$ $$leq M + epsilon.$$ Is this proof correct? Thank you.






calculus real-analysis proof-verification inequality uniform-convergence







share|cite|improve this question













Assume that $f:,mathbb{R}rightarrowmathbb{R}$ is a continuous function. Assume also that exists a $M>0$ such that, for all $x in mathbb{R}$, we have $$left|fleft(xright)right|leq M.$$ Suppose that exists a sequence of function $f_n$ such that $f_n rightarrow f$ uniformly. I want to prove that, for all $x in mathbb{R}$, we have $$left|f_{n}left(xright)right|leq M$$ for every sufficiently large $n$.



My attempt. Since $f_n rightarrow f$ uniformly for all $epsilon>0$ exists a $N>0$ such that for all $x in mathbb{R}$ and $n geq N$ we have $$left|f_{n}left(xright)-fleft(xright)right|<epsilon.$$ So $$left|f_{n}left(xright)right|<left|fleft(xright)right|+left|f_{n}left(xright)-fleft(xright)right|$$ $$leq M + epsilon.$$ Is this proof correct? Thank you.






calculus real-analysis proof-verification inequality uniform-convergence




calculus real-analysis proof-verification inequality uniform-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 11:34









User

986




986








  • 1




    What if $ f equiv M$ and $f_n equiv M+1/n$ ?
    – nicomezi
    Nov 22 at 11:54














  • 1




    What if $ f equiv M$ and $f_n equiv M+1/n$ ?
    – nicomezi
    Nov 22 at 11:54








1




1




What if $ f equiv M$ and $f_n equiv M+1/n$ ?
– nicomezi
Nov 22 at 11:54




What if $ f equiv M$ and $f_n equiv M+1/n$ ?
– nicomezi
Nov 22 at 11:54










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Your considerations are correct, but they do not show that $|f_n(x)| le M$ for all $x$ and all sufficiently large $n.$



You can not prove, what you want to prove!



Exanple : $f(x)=frac{1}{1+x^2}$, then $max {|f(x)|: x in mathbb R}=1=f(0)$.



Now take $f_n(x):=f(x)+frac{1}{n}$. Then



$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.






share|cite|improve this answer





















  • Sir ,As in OP question he did not mentioned that M is lub something .It can large also .Why we can not choose M+1 so that his argument will worK @fred
    – Shubham
    Nov 22 at 14:56










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:39










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:40











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009026%2funiform-convergence-and-inequalities%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Your considerations are correct, but they do not show that $|f_n(x)| le M$ for all $x$ and all sufficiently large $n.$



You can not prove, what you want to prove!



Exanple : $f(x)=frac{1}{1+x^2}$, then $max {|f(x)|: x in mathbb R}=1=f(0)$.



Now take $f_n(x):=f(x)+frac{1}{n}$. Then



$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.






share|cite|improve this answer





















  • Sir ,As in OP question he did not mentioned that M is lub something .It can large also .Why we can not choose M+1 so that his argument will worK @fred
    – Shubham
    Nov 22 at 14:56










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:39










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:40















up vote
1
down vote



accepted










Your considerations are correct, but they do not show that $|f_n(x)| le M$ for all $x$ and all sufficiently large $n.$



You can not prove, what you want to prove!



Exanple : $f(x)=frac{1}{1+x^2}$, then $max {|f(x)|: x in mathbb R}=1=f(0)$.



Now take $f_n(x):=f(x)+frac{1}{n}$. Then



$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.






share|cite|improve this answer





















  • Sir ,As in OP question he did not mentioned that M is lub something .It can large also .Why we can not choose M+1 so that his argument will worK @fred
    – Shubham
    Nov 22 at 14:56










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:39










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:40













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Your considerations are correct, but they do not show that $|f_n(x)| le M$ for all $x$ and all sufficiently large $n.$



You can not prove, what you want to prove!



Exanple : $f(x)=frac{1}{1+x^2}$, then $max {|f(x)|: x in mathbb R}=1=f(0)$.



Now take $f_n(x):=f(x)+frac{1}{n}$. Then



$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.






share|cite|improve this answer












Your considerations are correct, but they do not show that $|f_n(x)| le M$ for all $x$ and all sufficiently large $n.$



You can not prove, what you want to prove!



Exanple : $f(x)=frac{1}{1+x^2}$, then $max {|f(x)|: x in mathbb R}=1=f(0)$.



Now take $f_n(x):=f(x)+frac{1}{n}$. Then



$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 12:03









Fred

43.6k1644




43.6k1644












  • Sir ,As in OP question he did not mentioned that M is lub something .It can large also .Why we can not choose M+1 so that his argument will worK @fred
    – Shubham
    Nov 22 at 14:56










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:39










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:40


















  • Sir ,As in OP question he did not mentioned that M is lub something .It can large also .Why we can not choose M+1 so that his argument will worK @fred
    – Shubham
    Nov 22 at 14:56










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:39










  • yes the arguments work with $M+1$, but not with $M $.
    – Fred
    Nov 22 at 15:40
















Sir ,As in OP question he did not mentioned that M is lub something .It can large also .Why we can not choose M+1 so that his argument will worK @fred
– Shubham
Nov 22 at 14:56




Sir ,As in OP question he did not mentioned that M is lub something .It can large also .Why we can not choose M+1 so that his argument will worK @fred
– Shubham
Nov 22 at 14:56












yes the arguments work with $M+1$, but not with $M $.
– Fred
Nov 22 at 15:39




yes the arguments work with $M+1$, but not with $M $.
– Fred
Nov 22 at 15:39












yes the arguments work with $M+1$, but not with $M $.
– Fred
Nov 22 at 15:40




yes the arguments work with $M+1$, but not with $M $.
– Fred
Nov 22 at 15:40


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009026%2funiform-convergence-and-inequalities%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more