if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$
$begingroup$
Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$
Note: I don't have uniform convergence at my disposal.
Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$
$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.
I'm not sure where to use the supremum at I guess is my issue.
real-analysis integration
edited Jan 1 at 16:01
A.Γ.
22.8k32656
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
$endgroup$
add a comment |
$begingroup$
Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$
Note: I don't have uniform convergence at my disposal.
Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$
$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.
I'm not sure where to use the supremum at I guess is my issue.
real-analysis integration
edited Jan 1 at 16:01
A.Γ.
22.8k32656
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
$endgroup$
3
$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34
$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36
1
$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37
$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39
add a comment |
$begingroup$
Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$
Note: I don't have uniform convergence at my disposal.
Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$
$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.
I'm not sure where to use the supremum at I guess is my issue.
real-analysis integration
edited Jan 1 at 16:01
A.Γ.
22.8k32656
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
$endgroup$
Let $f, f_n : [a,b] to mathbb{R} $ be Riemann $forall n in mathbb{N}$ if $lim_{n to infty} sup _x |f_n(x) - f(x)| = 0$, then $lim _{n to infty} int_a^b f_n = int_a^b f$
Note: I don't have uniform convergence at my disposal.
Let $ varepsilon > 0$ there exists $N in mathbb{N}$ such that $forall n ge N$ $sup_{x in [a,b]} lvert f_n - f rvert < varepsilon$ Since $forall n, f_n, f$ are Riemann-Integrable which implies they are bounded. Let $M = max{| sup f_n|, : n in mathbb{N}} cup {|sup f|}$
$lvert int_a^b f_n - f rvert le int_a^b lvert f_n-f| le 2M(b-a)$.
I'm not sure where to use the supremum at I guess is my issue.
real-analysis integration
real-analysis integration
edited Jan 1 at 16:01
A.Γ.
22.8k32656
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
edited Jan 1 at 16:01
A.Γ.
22.8k32656
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
edited Jan 1 at 16:01
A.Γ.
22.8k32656
edited Jan 1 at 16:01
A.Γ.
22.8k32656
edited Jan 1 at 16:01
A.Γ.
22.8k32656
22.8k32656
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
asked Jan 31 '17 at 4:29
oliverjonesoliverjones
1,444932
1,444932
3
$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34
$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36
1
$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37
$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39
add a comment |
3
$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34
$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36
1
$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37
$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39
3
3
$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34
$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34
$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36
$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36
1
1
$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37
$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37
$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39
$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider $epsilon>0$
Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.
Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$
Then:
$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
$endgroup$
$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42
1
$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44
$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46
1
$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46
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',
autoActivateHeartbeat: false,
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%2f2121942%2fif-lim-n-to-infty-sup-x-f-nx-fx-0-then-lim-n-to-infty%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
$begingroup$
Consider $epsilon>0$
Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.
Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$
Then:
$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
$endgroup$
$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42
1
$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44
$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46
1
$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46
add a comment |
$begingroup$
Consider $epsilon>0$
Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.
Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$
Then:
$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
$endgroup$
$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42
1
$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44
$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46
1
$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46
add a comment |
$begingroup$
Consider $epsilon>0$
Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.
Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$
Then:
$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
$endgroup$
Consider $epsilon>0$
Then from $lim_{ntoinfty}sup_x|f_n(x)-f(x)|=0$ you can find a $Ninmathbb N$ such that $sup_x|f_n(x)-f(x)|<frac{epsilon}{b-a}$, for all $nge N$.
Which means $|f_n(x)-f(x)|<frac{epsilon}{b-a} forall xin[a,b] forall ngemathbb N$
Then:
$$left|int_a^b f_n-int_a^bfright|le int_a^b |f_n-f|lefrac{epsilon}{b-a}(b-a)=epsilon$$
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
answered Jan 31 '17 at 4:40
MomoMomo
12k21430
12k21430
$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42
1
$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44
$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46
1
$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46
add a comment |
$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42
1
$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44
$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46
1
$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46
$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42
$begingroup$
if $sup_x lvert f_n(x) - f(x) rvert < frac{varepsilon}{b-a} $ then $lvert f_n(x) - f(x) rvert< frac{varepsilon}{b-a} $ for all non-supremum values? cause the supremum gives me the largest difference? (Sorry, sometimes I cannot @ the person in my replies, must be my browser)
$endgroup$
– oliverjones
Jan 31 '17 at 4:42
1
1
$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44
$begingroup$
All elements of a set are smaller than its supremum, which is $xle sup A$ for all $xin A$
$endgroup$
– Momo
Jan 31 '17 at 4:44
$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46
$begingroup$
In other words, $lvert f_n(x) - f(x) rvert le sup_x lvert f_n(x) - f(x) rvert$ ?
$endgroup$
– oliverjones
Jan 31 '17 at 4:46
1
1
$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46
$begingroup$
Yes, that's right.
$endgroup$
– Momo
Jan 31 '17 at 4:46
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.
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%2f2121942%2fif-lim-n-to-infty-sup-x-f-nx-fx-0-then-lim-n-to-infty%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
3
$begingroup$
The hypothesis is that the convergence is uniform, so it is unclear what you mean by "I don't have uniform convergence at my disposal." I recommend using $int_a^b g(x),dxleq (b-a)suplimits_{aleq xleq b}g(x)$.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:34
$begingroup$
@JonasMeyer that is why I put that note because we haven't covered what uniform convergence is so I cannot yet use that in this proof even though I know that the hypothesis gives that it is uniform convergent.
$endgroup$
– oliverjones
Jan 31 '17 at 4:36
1
$begingroup$
I don't understand what you mean by "I cannot yet use that". You have to use it, even if you don't call it that. It is just a definition, and your hypothesis is that the definition is satisfied. Perhaps you mean you can't use extra theorems about uniform convergence? No need, you can use the given property (i.e. the uniform convergence) directly.
$endgroup$
– Jonas Meyer
Jan 31 '17 at 4:37
$begingroup$
@JonasMeyer yeah thats exactly what I mean. Sorry for the confusion.
$endgroup$
– oliverjones
Jan 31 '17 at 4:39