$A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$
$begingroup$
Let $X$ be a TVS ,and $A subset X$.
Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.
If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.
I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.
Any ideas?
Thanks for helping!
functional-analysis
asked Jan 4 at 21:46
user123user123
1,405316
$endgroup$
add a comment |
$begingroup$
Let $X$ be a TVS ,and $A subset X$.
Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.
If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.
I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.
Any ideas?
Thanks for helping!
functional-analysis
asked Jan 4 at 21:46
user123user123
1,405316
$endgroup$
add a comment |
$begingroup$
Let $X$ be a TVS ,and $A subset X$.
Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.
If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.
I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.
Any ideas?
Thanks for helping!
functional-analysis
asked Jan 4 at 21:46
user123user123
1,405316
$endgroup$
Let $X$ be a TVS ,and $A subset X$.
Then $A$ is convex $iff$ for all $s,t gt 0$ , $(s+t)A = sA + tA$.
If for all $s,tgt 0$ we have $(s+t)A = sA+tA$ then of course $A$ is convex.
I'm not sure about the other direction. It is always true that $(s+t)A subset sA+tA$ ,so assuming $A$ is convex I need to show the other inclusion.
Any ideas?
Thanks for helping!
functional-analysis
functional-analysis
asked Jan 4 at 21:46
user123user123
1,405316
asked Jan 4 at 21:46
user123user123
1,405316
asked Jan 4 at 21:46
user123user123
1,405316
asked Jan 4 at 21:46
user123user123
1,405316
asked Jan 4 at 21:46
user123user123
1,405316
1,405316
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
$endgroup$
$begingroup$
So easy. Thanks!
$endgroup$
– user123
Jan 4 at 21:56
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
$endgroup$
$begingroup$
So easy. Thanks!
$endgroup$
– user123
Jan 4 at 21:56
add a comment |
$begingroup$
Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
$endgroup$
$begingroup$
So easy. Thanks!
$endgroup$
– user123
Jan 4 at 21:56
add a comment |
$begingroup$
Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
$endgroup$
Let $sx+tyin sA+tA$. Then $$(s+t)left(frac{s}{s+t}x+frac{t}{s+t}yright)in(s+t)A$$
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
answered Jan 4 at 21:53
SmileyCraftSmileyCraft
3,739519
3,739519
$begingroup$
So easy. Thanks!
$endgroup$
– user123
Jan 4 at 21:56
add a comment |
$begingroup$
So easy. Thanks!
$endgroup$
– user123
Jan 4 at 21:56
$begingroup$
So easy. Thanks!
$endgroup$
– user123
Jan 4 at 21:56
$begingroup$
So easy. Thanks!
$endgroup$
– user123
Jan 4 at 21:56
add a comment |
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