An interesting integral inequality

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
5
down vote

favorite
2












Anyone has an idea to solve the following problem?



Let $fin C^2left(left[-1,1right]right)$ such that $f''left(xright)geq0$
for any $xinleft[-1,1right]$ and $fleft(-1right)=fleft(1right)=0.$
Prove that
$$
max_left[-1,1right]left|fright|leq10left(intop_-1^1(f''left(xright))^2dxright)^frac12.
$$






real-analysis integral-inequality







share|cite|improve this question





















  • What does $fin C^2([-1,1])$ mean?
    – AnotherJohnDoe
    2 hours ago






  • 2




    It means: $f$ is continuously differentiable up to second order on a neighborhood of [-1,1].
    – Hahn
    2 hours ago















up vote
5
down vote

favorite
2












Anyone has an idea to solve the following problem?



Let $fin C^2left(left[-1,1right]right)$ such that $f''left(xright)geq0$
for any $xinleft[-1,1right]$ and $fleft(-1right)=fleft(1right)=0.$
Prove that
$$
max_left[-1,1right]left|fright|leq10left(intop_-1^1(f''left(xright))^2dxright)^frac12.
$$






real-analysis integral-inequality







share|cite|improve this question





















  • What does $fin C^2([-1,1])$ mean?
    – AnotherJohnDoe
    2 hours ago






  • 2




    It means: $f$ is continuously differentiable up to second order on a neighborhood of [-1,1].
    – Hahn
    2 hours ago













up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2





Anyone has an idea to solve the following problem?



Let $fin C^2left(left[-1,1right]right)$ such that $f''left(xright)geq0$
for any $xinleft[-1,1right]$ and $fleft(-1right)=fleft(1right)=0.$
Prove that
$$
max_left[-1,1right]left|fright|leq10left(intop_-1^1(f''left(xright))^2dxright)^frac12.
$$






real-analysis integral-inequality







share|cite|improve this question













Anyone has an idea to solve the following problem?



Let $fin C^2left(left[-1,1right]right)$ such that $f''left(xright)geq0$
for any $xinleft[-1,1right]$ and $fleft(-1right)=fleft(1right)=0.$
Prove that
$$
max_left[-1,1right]left|fright|leq10left(intop_-1^1(f''left(xright))^2dxright)^frac12.
$$






real-analysis integral-inequality




real-analysis integral-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









Hahn

663




663











  • What does $fin C^2([-1,1])$ mean?
    – AnotherJohnDoe
    2 hours ago






  • 2




    It means: $f$ is continuously differentiable up to second order on a neighborhood of [-1,1].
    – Hahn
    2 hours ago

















  • What does $fin C^2([-1,1])$ mean?
    – AnotherJohnDoe
    2 hours ago






  • 2




    It means: $f$ is continuously differentiable up to second order on a neighborhood of [-1,1].
    – Hahn
    2 hours ago
















What does $fin C^2([-1,1])$ mean?
– AnotherJohnDoe
2 hours ago




What does $fin C^2([-1,1])$ mean?
– AnotherJohnDoe
2 hours ago




2




2




It means: $f$ is continuously differentiable up to second order on a neighborhood of [-1,1].
– Hahn
2 hours ago





It means: $f$ is continuously differentiable up to second order on a neighborhood of [-1,1].
– Hahn
2 hours ago











1 Answer
1






active

oldest

votes

















up vote
6
down vote













Let $alpha$ be a maximum point of $|f|$. Then $f'(alpha) = 0$ and $f(alpha) leq 0$. So



$$ max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx. $$



Now by the Cauchy-Schwarz inequality,



beginalign*
int_alpha^1 (1-x) f''(x) , dx
&leq left( int_alpha^1 (1-x)^2 , dx right)^1/2left( int_-1^1 f''(x)^2 , dx right)^1/2 \
&leq sqrtfrac83 left( int_-1^1 f''(x)^2 , dx right)^1/2.
endalign*



Remarks. Let $C_0$ be the infimum over all $C>0$ for which



$$max_[-1,1] |f| leq C left( int_-1^1 f''(x)^2 , dx right)^1/2$$



holds for all $f in C^2([-1,1])$ with $f(-1) = f(1) = 0$. The above proof shows that $C_0 leq sqrt8/3$. In fact, a slightly better idea yields an improved bound $C_0 leq sqrt2/3$. It would be interesting to identify the sharp constant $C_0$.



Moreover, the convexity condition $f'' geq 0$ is not essential to the above solution. It would be nice if the sharp bound $C_0$ can be improved if we restrict our attention to convex functions.






share|cite|improve this answer






















  • It's such a nice transform. $$max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx.$$
    – Hahn
    1 hour ago











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: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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%2f2953643%2fan-interesting-integral-inequality%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote













Let $alpha$ be a maximum point of $|f|$. Then $f'(alpha) = 0$ and $f(alpha) leq 0$. So



$$ max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx. $$



Now by the Cauchy-Schwarz inequality,



beginalign*
int_alpha^1 (1-x) f''(x) , dx
&leq left( int_alpha^1 (1-x)^2 , dx right)^1/2left( int_-1^1 f''(x)^2 , dx right)^1/2 \
&leq sqrtfrac83 left( int_-1^1 f''(x)^2 , dx right)^1/2.
endalign*



Remarks. Let $C_0$ be the infimum over all $C>0$ for which



$$max_[-1,1] |f| leq C left( int_-1^1 f''(x)^2 , dx right)^1/2$$



holds for all $f in C^2([-1,1])$ with $f(-1) = f(1) = 0$. The above proof shows that $C_0 leq sqrt8/3$. In fact, a slightly better idea yields an improved bound $C_0 leq sqrt2/3$. It would be interesting to identify the sharp constant $C_0$.



Moreover, the convexity condition $f'' geq 0$ is not essential to the above solution. It would be nice if the sharp bound $C_0$ can be improved if we restrict our attention to convex functions.






share|cite|improve this answer






















  • It's such a nice transform. $$max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx.$$
    – Hahn
    1 hour ago















up vote
6
down vote













Let $alpha$ be a maximum point of $|f|$. Then $f'(alpha) = 0$ and $f(alpha) leq 0$. So



$$ max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx. $$



Now by the Cauchy-Schwarz inequality,



beginalign*
int_alpha^1 (1-x) f''(x) , dx
&leq left( int_alpha^1 (1-x)^2 , dx right)^1/2left( int_-1^1 f''(x)^2 , dx right)^1/2 \
&leq sqrtfrac83 left( int_-1^1 f''(x)^2 , dx right)^1/2.
endalign*



Remarks. Let $C_0$ be the infimum over all $C>0$ for which



$$max_[-1,1] |f| leq C left( int_-1^1 f''(x)^2 , dx right)^1/2$$



holds for all $f in C^2([-1,1])$ with $f(-1) = f(1) = 0$. The above proof shows that $C_0 leq sqrt8/3$. In fact, a slightly better idea yields an improved bound $C_0 leq sqrt2/3$. It would be interesting to identify the sharp constant $C_0$.



Moreover, the convexity condition $f'' geq 0$ is not essential to the above solution. It would be nice if the sharp bound $C_0$ can be improved if we restrict our attention to convex functions.






share|cite|improve this answer






















  • It's such a nice transform. $$max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx.$$
    – Hahn
    1 hour ago













up vote
6
down vote










up vote
6
down vote









Let $alpha$ be a maximum point of $|f|$. Then $f'(alpha) = 0$ and $f(alpha) leq 0$. So



$$ max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx. $$



Now by the Cauchy-Schwarz inequality,



beginalign*
int_alpha^1 (1-x) f''(x) , dx
&leq left( int_alpha^1 (1-x)^2 , dx right)^1/2left( int_-1^1 f''(x)^2 , dx right)^1/2 \
&leq sqrtfrac83 left( int_-1^1 f''(x)^2 , dx right)^1/2.
endalign*



Remarks. Let $C_0$ be the infimum over all $C>0$ for which



$$max_[-1,1] |f| leq C left( int_-1^1 f''(x)^2 , dx right)^1/2$$



holds for all $f in C^2([-1,1])$ with $f(-1) = f(1) = 0$. The above proof shows that $C_0 leq sqrt8/3$. In fact, a slightly better idea yields an improved bound $C_0 leq sqrt2/3$. It would be interesting to identify the sharp constant $C_0$.



Moreover, the convexity condition $f'' geq 0$ is not essential to the above solution. It would be nice if the sharp bound $C_0$ can be improved if we restrict our attention to convex functions.






share|cite|improve this answer














Let $alpha$ be a maximum point of $|f|$. Then $f'(alpha) = 0$ and $f(alpha) leq 0$. So



$$ max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx. $$



Now by the Cauchy-Schwarz inequality,



beginalign*
int_alpha^1 (1-x) f''(x) , dx
&leq left( int_alpha^1 (1-x)^2 , dx right)^1/2left( int_-1^1 f''(x)^2 , dx right)^1/2 \
&leq sqrtfrac83 left( int_-1^1 f''(x)^2 , dx right)^1/2.
endalign*



Remarks. Let $C_0$ be the infimum over all $C>0$ for which



$$max_[-1,1] |f| leq C left( int_-1^1 f''(x)^2 , dx right)^1/2$$



holds for all $f in C^2([-1,1])$ with $f(-1) = f(1) = 0$. The above proof shows that $C_0 leq sqrt8/3$. In fact, a slightly better idea yields an improved bound $C_0 leq sqrt2/3$. It would be interesting to identify the sharp constant $C_0$.



Moreover, the convexity condition $f'' geq 0$ is not essential to the above solution. It would be nice if the sharp bound $C_0$ can be improved if we restrict our attention to convex functions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









Sangchul Lee

88.1k12159260




88.1k12159260











  • It's such a nice transform. $$max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx.$$
    – Hahn
    1 hour ago

















  • It's such a nice transform. $$max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx.$$
    – Hahn
    1 hour ago
















It's such a nice transform. $$max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx.$$
– Hahn
1 hour ago





It's such a nice transform. $$max_[-1,1] |f| = -f(alpha) = int_alpha^1 f'(x) , dx = int_alpha^1 (1-x) f''(x) , dx.$$
– Hahn
1 hour ago


















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2953643%2fan-interesting-integral-inequality%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What does second last employer means? [closed]

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
Read more

List of Gilmore Girls characters

Image
The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
Read more

Confectionery

Image
Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of ...
Read more