How many Convex and Odd functions exist?
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During lunch this question popped on mind.
How many odd and convex functions $f: R to R $ exist ?
I guessed just 1, namely $f(x)=ax$ where $a$ is a fixed real number. But I couldn't either prove it or provide counter example.
real-analysis complex-analysis
asked Nov 13 at 22:54
Red shoes
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up vote
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During lunch this question popped on mind.
How many odd and convex functions $f: R to R $ exist ?
I guessed just 1, namely $f(x)=ax$ where $a$ is a fixed real number. But I couldn't either prove it or provide counter example.
real-analysis complex-analysis
asked Nov 13 at 22:54
Red shoes
4,556621
@user376343 Convexity implies continuity.
– T. Bongers
Nov 13 at 23:04
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
During lunch this question popped on mind.
How many odd and convex functions $f: R to R $ exist ?
I guessed just 1, namely $f(x)=ax$ where $a$ is a fixed real number. But I couldn't either prove it or provide counter example.
real-analysis complex-analysis
asked Nov 13 at 22:54
Red shoes
4,556621
During lunch this question popped on mind.
How many odd and convex functions $f: R to R $ exist ?
I guessed just 1, namely $f(x)=ax$ where $a$ is a fixed real number. But I couldn't either prove it or provide counter example.
real-analysis complex-analysis
real-analysis complex-analysis
asked Nov 13 at 22:54
Red shoes
4,556621
asked Nov 13 at 22:54
Red shoes
4,556621
asked Nov 13 at 22:54
Red shoes
4,556621
asked Nov 13 at 22:54
Red shoes
4,556621
asked Nov 13 at 22:54
Red shoes
4,556621
4,556621
@user376343 Convexity implies continuity.
– T. Bongers
Nov 13 at 23:04
add a comment |
@user376343 Convexity implies continuity.
– T. Bongers
Nov 13 at 23:04
@user376343 Convexity implies continuity.
– T. Bongers
Nov 13 at 23:04
@user376343 Convexity implies continuity.
– T. Bongers
Nov 13 at 23:04
add a comment |
1 Answer
1
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up vote
3
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accepted
Scalar multiples of $x$ are the only examples.
To see why, suppose you're given $(1, f(1))$ as well as $(0, 0) = (0, f(0))$. Draw the line connecting these two points, as well as its extension through $(-1, f(-1))$.
Since $f$ is convex, its graph for $0 < x < 1$ lies on or below this line. If the graph ever dips below the line, then using the oddness of $f$ implies that it gets above the line connecting $-1$ and $0$, contradicting convexity.
Hence, $f$ must agree with a straight line for $x in (0, 1)$. It's not hard to extend this argument to $mathbb{R}$.
answered Nov 13 at 22:59
T. Bongers
22.2k54359
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Scalar multiples of $x$ are the only examples.
To see why, suppose you're given $(1, f(1))$ as well as $(0, 0) = (0, f(0))$. Draw the line connecting these two points, as well as its extension through $(-1, f(-1))$.
Since $f$ is convex, its graph for $0 < x < 1$ lies on or below this line. If the graph ever dips below the line, then using the oddness of $f$ implies that it gets above the line connecting $-1$ and $0$, contradicting convexity.
Hence, $f$ must agree with a straight line for $x in (0, 1)$. It's not hard to extend this argument to $mathbb{R}$.
answered Nov 13 at 22:59
T. Bongers
22.2k54359
add a comment |
up vote
3
down vote
accepted
Scalar multiples of $x$ are the only examples.
To see why, suppose you're given $(1, f(1))$ as well as $(0, 0) = (0, f(0))$. Draw the line connecting these two points, as well as its extension through $(-1, f(-1))$.
Since $f$ is convex, its graph for $0 < x < 1$ lies on or below this line. If the graph ever dips below the line, then using the oddness of $f$ implies that it gets above the line connecting $-1$ and $0$, contradicting convexity.
Hence, $f$ must agree with a straight line for $x in (0, 1)$. It's not hard to extend this argument to $mathbb{R}$.
answered Nov 13 at 22:59
T. Bongers
22.2k54359
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Scalar multiples of $x$ are the only examples.
To see why, suppose you're given $(1, f(1))$ as well as $(0, 0) = (0, f(0))$. Draw the line connecting these two points, as well as its extension through $(-1, f(-1))$.
Since $f$ is convex, its graph for $0 < x < 1$ lies on or below this line. If the graph ever dips below the line, then using the oddness of $f$ implies that it gets above the line connecting $-1$ and $0$, contradicting convexity.
Hence, $f$ must agree with a straight line for $x in (0, 1)$. It's not hard to extend this argument to $mathbb{R}$.
answered Nov 13 at 22:59
T. Bongers
22.2k54359
Scalar multiples of $x$ are the only examples.
To see why, suppose you're given $(1, f(1))$ as well as $(0, 0) = (0, f(0))$. Draw the line connecting these two points, as well as its extension through $(-1, f(-1))$.
Since $f$ is convex, its graph for $0 < x < 1$ lies on or below this line. If the graph ever dips below the line, then using the oddness of $f$ implies that it gets above the line connecting $-1$ and $0$, contradicting convexity.
Hence, $f$ must agree with a straight line for $x in (0, 1)$. It's not hard to extend this argument to $mathbb{R}$.
answered Nov 13 at 22:59
T. Bongers
22.2k54359
answered Nov 13 at 22:59
T. Bongers
22.2k54359
answered Nov 13 at 22:59
T. Bongers
22.2k54359
answered Nov 13 at 22:59
T. Bongers
22.2k54359
22.2k54359
add a comment |
add a comment |
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@user376343 Convexity implies continuity.
– T. Bongers
Nov 13 at 23:04