Otdfbt

Why did the Death Eaters wait to reopen the Chamber of Secrets?

Must a CPU have a GPU if the motherboard provides a display port (when there isn't any separate video card)?

ifnum expanding too much - what is happening?

Is it possible to have battery technology that can't be duplicated?

Can a 40amp breaker be used safely and without issue with a 40amp device on 6AWG wire?

What did the 8086 (and 8088) do upon encountering an illegal instruction?

Print "N NE E SE S SW W NW"

Harley Davidson clattering noise from engine, backfire and failure to start

How can I find out about the game world without meta-influencing it?

Why did the AvroCar fail to fly above 3 feet?

ISP is not hashing the password I log in with online. Should I take any action?

How to represent jealousy in a cute way?

Why does this Apple //e drops into system monitor when booting?

Fastest way from 10 to 1 with everyone in between

Boss making me feel guilty for leaving the company at the end of my internship

Parsing text written the millitext font

Dedicated bike GPS computer over smartphone

Is pointing finger in meeting consider bad?

Idiom for 'person who gets violent when drunk"

How effective would a full set of plate armor be against wild animals found in temperate regions (bears, snakes, wolves)?

Any gotchas in buying second-hand sanitary ware?

Parallelized for loop in Bash

Would a character with eternal youth be AL compliant?

Is there a term for when fiction refers to fiction

Minimum point-wise of convex and scalar

Proving that the maximum of two convex functions is also convexProving that $|x|^p,p geq 1$ is convexConvex functions - two questionsLogarithmically Convex FunctionProve local minimum of a convex function is a global minumum (using only convexity)How to determine whether a function is concave, convex, quasi-concave and quasi-convexIs the following set convex?Convexity of product of two convex functionsConvexity of distance function to a point and related propertiesProof (without use of differential calculus) that $e^sqrtx$ is convex on $[1,+infty)$.Without use of derivatives, prove that function $(1+x^p)^1/p$ is convex for $pgeq 1$

2

$begingroup$

Let $ f(x) = log_3 (1+5^x) $ and $ k geq 0 $. I need to prove that $ frac1min f(x), k $ is convex.

My attempt:

Since $ 1+5^x geq 1$, then $ log_3 (1+5^x) geq 0 $. Furthermore, since $ 1+5^x geq 1$ is monotonically increasing, $ log_3 (1+5^x) $ all also be so. Thus,

$$
min log_3 (1+5^x), k rightarrow min 1+5^x, 3^k
$$

So, now I can equivalently try to prove $ min 1+5^x, 3^k $ that looks simpler. I have tried using the convexity definition, but I have not been successful.

optimization convex-analysis convex-optimization

share|cite|improve this question

edited May 29 at 10:22

Dunkel

asked May 29 at 10:00

DunkelDunkel

484312

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean $max$? If not I don't think it is convex.
  $endgroup$
  – Coolwater
  May 29 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant $ 1/ min(.)$
  $endgroup$
  – Dunkel
  May 29 at 10:22
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $ f(x) = log_3 (1+5^x) $ and $ k geq 0 $. I need to prove that $ frac1min f(x), k $ is convex.

My attempt:

Since $ 1+5^x geq 1$, then $ log_3 (1+5^x) geq 0 $. Furthermore, since $ 1+5^x geq 1$ is monotonically increasing, $ log_3 (1+5^x) $ all also be so. Thus,

$$
min log_3 (1+5^x), k rightarrow min 1+5^x, 3^k
$$

So, now I can equivalently try to prove $ min 1+5^x, 3^k $ that looks simpler. I have tried using the convexity definition, but I have not been successful.

optimization convex-analysis convex-optimization

share|cite|improve this question

edited May 29 at 10:22

Dunkel

asked May 29 at 10:00

DunkelDunkel

484312

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean $max$? If not I don't think it is convex.
  $endgroup$
  – Coolwater
  May 29 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant $ 1/ min(.)$
  $endgroup$
  – Dunkel
  May 29 at 10:22
  

  
  
  
  

add a comment | 

$begingroup$

Let $ f(x) = log_3 (1+5^x) $ and $ k geq 0 $. I need to prove that $ frac1min f(x), k $ is convex.

My attempt:

Since $ 1+5^x geq 1$, then $ log_3 (1+5^x) geq 0 $. Furthermore, since $ 1+5^x geq 1$ is monotonically increasing, $ log_3 (1+5^x) $ all also be so. Thus,

$$
min log_3 (1+5^x), k rightarrow min 1+5^x, 3^k
$$

So, now I can equivalently try to prove $ min 1+5^x, 3^k $ that looks simpler. I have tried using the convexity definition, but I have not been successful.

optimization convex-analysis convex-optimization

share|cite|improve this question

edited May 29 at 10:22

Dunkel

asked May 29 at 10:00

DunkelDunkel

484312

$endgroup$

share|cite|improve this question

edited May 29 at 10:22

Dunkel

asked May 29 at 10:00

DunkelDunkel

484312

share|cite|improve this question

edited May 29 at 10:22

Dunkel

asked May 29 at 10:00

DunkelDunkel

484312

asked May 29 at 10:00

DunkelDunkel

484312

asked May 29 at 10:00

DunkelDunkel

484312

4 Answers
4

active

oldest

votes

2

$begingroup$

If you rewrite
$$
frac1minf(x),k=maxleftfrac1f(x),frac1kright
$$
then what is left is to prove that
$$
frac1f(x)=frac1log_3(1+5^x)=fracln 3ln(1+5^x)=fracln 3ln(1+e^xln 5)
$$
is convex (as the max of two convex functions is convex). It can be done by proving first that the function
$$
g(t)=frac1ln(1+e^t)
$$
is convex (check that $g''(t)ge 0$), then
$$
frac1f(x)=ln 3cdot g(xln 5)
$$
is convex too.

share|cite|improve this answer

answered May 29 at 15:53

A.Γ.A.Γ.

23.3k32758

$endgroup$

add a comment | 

2

$begingroup$

Plot for

$$
g(x,k)=frac1min[k,log_3(1+5^x)]
$$

with $k = 1$

share|cite|improve this answer

answered May 29 at 14:09

CesareoCesareo

11.2k3520

$endgroup$

add a comment | 

1

$begingroup$

I believe you mean $max$, otherwise the function is not convex.

Suppose that $max f(x), k = f(x)$ for $x geq a$, and $k$ otherwise. The definition of convexity states that the function is convex if every chord lies above the function. It is easy to show that $f(x)$ is convex, by differentiating twice. So every chord on the range $[a,infty)$ lies above the function. Clearly any chord on $(-infty, a]$ is simply horizontal and therefore lies under the function. Now consider any chord with endpoints at $x=b, c$ where $b<a<c$. From drawing the graph and noting that $f(x)$ is increasing, it is easy to see that this chord lies above the chord with endpoints $x=a, c$, and this chord in turn lies above the function on that range. Furthermore, this chord clearly lies above $x=k$ over the range $[b,c]$. So combining these observations, the chord lies above the function, and we are done.

share|cite|improve this answer

answered May 29 at 10:09

auscryptauscrypt

7,306614

$endgroup$

add a comment | 

1

$begingroup$

$mink,log_3(1+5^x)$ is not convex, unless $k=0$.
However, $maxk,log_3(1+5^x)$ is, because the maximum of two convex functions is convex. For a proof, see here.

share|cite|improve this answer

answered May 29 at 10:12

uniquesolutionuniquesolution

10.9k1825

$endgroup$

add a comment | 

Your Answer

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3243874%2fminimum-point-wise-of-convex-and-scalar%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

If you rewrite
$$
frac1minf(x),k=maxleftfrac1f(x),frac1kright
$$
then what is left is to prove that
$$
frac1f(x)=frac1log_3(1+5^x)=fracln 3ln(1+5^x)=fracln 3ln(1+e^xln 5)
$$
is convex (as the max of two convex functions is convex). It can be done by proving first that the function
$$
g(t)=frac1ln(1+e^t)
$$
is convex (check that $g''(t)ge 0$), then
$$
frac1f(x)=ln 3cdot g(xln 5)
$$
is convex too.

share|cite|improve this answer

answered May 29 at 15:53

A.Γ.A.Γ.

23.3k32758

$endgroup$

add a comment | 

2

$begingroup$

If you rewrite
$$
frac1minf(x),k=maxleftfrac1f(x),frac1kright
$$
then what is left is to prove that
$$
frac1f(x)=frac1log_3(1+5^x)=fracln 3ln(1+5^x)=fracln 3ln(1+e^xln 5)
$$
is convex (as the max of two convex functions is convex). It can be done by proving first that the function
$$
g(t)=frac1ln(1+e^t)
$$
is convex (check that $g''(t)ge 0$), then
$$
frac1f(x)=ln 3cdot g(xln 5)
$$
is convex too.

share|cite|improve this answer

answered May 29 at 15:53

A.Γ.A.Γ.

23.3k32758

$endgroup$

add a comment | 

$begingroup$

If you rewrite
$$
frac1minf(x),k=maxleftfrac1f(x),frac1kright
$$
then what is left is to prove that
$$
frac1f(x)=frac1log_3(1+5^x)=fracln 3ln(1+5^x)=fracln 3ln(1+e^xln 5)
$$
is convex (as the max of two convex functions is convex). It can be done by proving first that the function
$$
g(t)=frac1ln(1+e^t)
$$
is convex (check that $g''(t)ge 0$), then
$$
frac1f(x)=ln 3cdot g(xln 5)
$$
is convex too.

share|cite|improve this answer

answered May 29 at 15:53

A.Γ.A.Γ.

23.3k32758

$endgroup$

share|cite|improve this answer

answered May 29 at 15:53

A.Γ.A.Γ.

23.3k32758

answered May 29 at 15:53

A.Γ.A.Γ.

23.3k32758

answered May 29 at 15:53

A.Γ.A.Γ.

23.3k32758

2

$begingroup$

Plot for

$$
g(x,k)=frac1min[k,log_3(1+5^x)]
$$

with $k = 1$

share|cite|improve this answer

answered May 29 at 14:09

CesareoCesareo

11.2k3520

$endgroup$

add a comment | 

2

$begingroup$

Plot for

$$
g(x,k)=frac1min[k,log_3(1+5^x)]
$$

with $k = 1$

share|cite|improve this answer

answered May 29 at 14:09

CesareoCesareo

11.2k3520

$endgroup$

add a comment | 

$begingroup$

Plot for

$$
g(x,k)=frac1min[k,log_3(1+5^x)]
$$

with $k = 1$

share|cite|improve this answer

answered May 29 at 14:09

CesareoCesareo

11.2k3520

$endgroup$

share|cite|improve this answer

answered May 29 at 14:09

CesareoCesareo

11.2k3520

answered May 29 at 14:09

CesareoCesareo

11.2k3520

answered May 29 at 14:09

CesareoCesareo

11.2k3520

1

$begingroup$

I believe you mean $max$, otherwise the function is not convex.

Suppose that $max f(x), k = f(x)$ for $x geq a$, and $k$ otherwise. The definition of convexity states that the function is convex if every chord lies above the function. It is easy to show that $f(x)$ is convex, by differentiating twice. So every chord on the range $[a,infty)$ lies above the function. Clearly any chord on $(-infty, a]$ is simply horizontal and therefore lies under the function. Now consider any chord with endpoints at $x=b, c$ where $b<a<c$. From drawing the graph and noting that $f(x)$ is increasing, it is easy to see that this chord lies above the chord with endpoints $x=a, c$, and this chord in turn lies above the function on that range. Furthermore, this chord clearly lies above $x=k$ over the range $[b,c]$. So combining these observations, the chord lies above the function, and we are done.

share|cite|improve this answer

answered May 29 at 10:09

auscryptauscrypt

7,306614

$endgroup$

add a comment | 

1

$begingroup$

I believe you mean $max$, otherwise the function is not convex.

Suppose that $max f(x), k = f(x)$ for $x geq a$, and $k$ otherwise. The definition of convexity states that the function is convex if every chord lies above the function. It is easy to show that $f(x)$ is convex, by differentiating twice. So every chord on the range $[a,infty)$ lies above the function. Clearly any chord on $(-infty, a]$ is simply horizontal and therefore lies under the function. Now consider any chord with endpoints at $x=b, c$ where $b<a<c$. From drawing the graph and noting that $f(x)$ is increasing, it is easy to see that this chord lies above the chord with endpoints $x=a, c$, and this chord in turn lies above the function on that range. Furthermore, this chord clearly lies above $x=k$ over the range $[b,c]$. So combining these observations, the chord lies above the function, and we are done.

share|cite|improve this answer

answered May 29 at 10:09

auscryptauscrypt

7,306614

$endgroup$

add a comment | 

$begingroup$

I believe you mean $max$, otherwise the function is not convex.

Suppose that $max f(x), k = f(x)$ for $x geq a$, and $k$ otherwise. The definition of convexity states that the function is convex if every chord lies above the function. It is easy to show that $f(x)$ is convex, by differentiating twice. So every chord on the range $[a,infty)$ lies above the function. Clearly any chord on $(-infty, a]$ is simply horizontal and therefore lies under the function. Now consider any chord with endpoints at $x=b, c$ where $b<a<c$. From drawing the graph and noting that $f(x)$ is increasing, it is easy to see that this chord lies above the chord with endpoints $x=a, c$, and this chord in turn lies above the function on that range. Furthermore, this chord clearly lies above $x=k$ over the range $[b,c]$. So combining these observations, the chord lies above the function, and we are done.

share|cite|improve this answer

answered May 29 at 10:09

auscryptauscrypt

7,306614

$endgroup$

share|cite|improve this answer

answered May 29 at 10:09

auscryptauscrypt

7,306614

answered May 29 at 10:09

auscryptauscrypt

7,306614

answered May 29 at 10:09

auscryptauscrypt

7,306614

1

$begingroup$

$mink,log_3(1+5^x)$ is not convex, unless $k=0$.
However, $maxk,log_3(1+5^x)$ is, because the maximum of two convex functions is convex. For a proof, see here.

share|cite|improve this answer

answered May 29 at 10:12

uniquesolutionuniquesolution

10.9k1825

$endgroup$

add a comment | 

1

$begingroup$

$mink,log_3(1+5^x)$ is not convex, unless $k=0$.
However, $maxk,log_3(1+5^x)$ is, because the maximum of two convex functions is convex. For a proof, see here.

share|cite|improve this answer

answered May 29 at 10:12

uniquesolutionuniquesolution

10.9k1825

$endgroup$

add a comment | 

$begingroup$

$mink,log_3(1+5^x)$ is not convex, unless $k=0$.
However, $maxk,log_3(1+5^x)$ is, because the maximum of two convex functions is convex. For a proof, see here.

share|cite|improve this answer

answered May 29 at 10:12

uniquesolutionuniquesolution

10.9k1825

$endgroup$

share|cite|improve this answer

answered May 29 at 10:12

uniquesolutionuniquesolution

10.9k1825

answered May 29 at 10:12

uniquesolutionuniquesolution

10.9k1825

answered May 29 at 10:12

uniquesolutionuniquesolution

10.9k1825