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
asked Nov 22 at 11:34
User
986
add a comment |
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
asked Nov 22 at 11:34
User
986
1
What if $ f equiv M$ and $f_n equiv M+1/n$ ?
– nicomezi
Nov 22 at 11:54
add a comment |
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
asked Nov 22 at 11:34
User
986
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
asked Nov 22 at 11:34
User
986
asked Nov 22 at 11:34
User
986
asked Nov 22 at 11:34
User
986
asked Nov 22 at 11:34
User
986
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
add a comment |
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
add a comment |
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
$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.
answered Nov 22 at 12:03
Fred
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 $.
– Fred
Nov 22 at 15:40
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.
answered Nov 22 at 12:03
Fred
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
add a comment |
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
$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.
answered Nov 22 at 12:03
Fred
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
add a comment |
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
$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.
answered Nov 22 at 12:03
Fred
43.6k1644
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
$max {|f_n(x)|: x in mathbb R}=1+frac{1}{n}=f_n(0)$.
answered Nov 22 at 12:03
Fred
43.6k1644
answered Nov 22 at 12:03
Fred
43.6k1644
answered Nov 22 at 12:03
Fred
43.6k1644
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
add a comment |
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
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
What if $ f equiv M$ and $f_n equiv M+1/n$ ?
– nicomezi
Nov 22 at 11:54