Prove that a convex function minus a constant is convexShow that the set of all minimum points of a convex function over a convex set is convexProve that a given function is convexCharacterization convex function.Square of a convex non-negative function is still convexProve local minimum of a convex function is a global minumum (using only convexity)Prove that function is convexJensen's inequality and convex functionsA function with convex level sets but is not a convex functionConvex conjugate of the composition of a norm with a convex functionEquivalence of convex function
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Prove that a convex function minus a constant is convex
Show that the set of all minimum points of a convex function over a convex set is convexProve that a given function is convexCharacterization convex function.Square of a convex non-negative function is still convexProve local minimum of a convex function is a global minumum (using only convexity)Prove that function is convexJensen's inequality and convex functionsA function with convex level sets but is not a convex functionConvex conjugate of the composition of a norm with a convex functionEquivalence of convex function
$begingroup$
Let $f : mathbbR^d rightarrow mathbbR$, be a convex function, so $forall x, y in mathbbR^d$
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)$$
$t in[0,1]$. How do you show that $s(x) = f(x)-k$ also is convex ($kinmathbbR$)?
I only get:
$$s(tx+(1-t)y) leq tf(x)+(1-t)f(y) -k =ts(x)+(1-t)s(y)+k$$
convex-analysis
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
asked Mar 17 at 22:31
User123456789User123456789
16928
$endgroup$
add a comment |
$begingroup$
Let $f : mathbbR^d rightarrow mathbbR$, be a convex function, so $forall x, y in mathbbR^d$
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)$$
$t in[0,1]$. How do you show that $s(x) = f(x)-k$ also is convex ($kinmathbbR$)?
I only get:
$$s(tx+(1-t)y) leq tf(x)+(1-t)f(y) -k =ts(x)+(1-t)s(y)+k$$
convex-analysis
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
asked Mar 17 at 22:31
User123456789User123456789
16928
$endgroup$
add a comment |
$begingroup$
Let $f : mathbbR^d rightarrow mathbbR$, be a convex function, so $forall x, y in mathbbR^d$
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)$$
$t in[0,1]$. How do you show that $s(x) = f(x)-k$ also is convex ($kinmathbbR$)?
I only get:
$$s(tx+(1-t)y) leq tf(x)+(1-t)f(y) -k =ts(x)+(1-t)s(y)+k$$
convex-analysis
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
asked Mar 17 at 22:31
User123456789User123456789
16928
$endgroup$
Let $f : mathbbR^d rightarrow mathbbR$, be a convex function, so $forall x, y in mathbbR^d$
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)$$
$t in[0,1]$. How do you show that $s(x) = f(x)-k$ also is convex ($kinmathbbR$)?
I only get:
$$s(tx+(1-t)y) leq tf(x)+(1-t)f(y) -k =ts(x)+(1-t)s(y)+k$$
convex-analysis
convex-analysis
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
asked Mar 17 at 22:31
User123456789User123456789
16928
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
asked Mar 17 at 22:31
User123456789User123456789
16928
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
edited Mar 17 at 22:53
J. W. Tanner
3,9221320
3,9221320
asked Mar 17 at 22:31
User123456789User123456789
16928
asked Mar 17 at 22:31
User123456789User123456789
16928
asked Mar 17 at 22:31
User123456789User123456789
16928
16928
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I think you just have an arithmetic error; namely,
$$s(tx + (1-t)y) leq tf(x) + (1-t)f(y) - k = t(f(x)-k) + (1-t)(f(y)-k).$$
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
$endgroup$
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you just have an arithmetic error; namely,
$$s(tx + (1-t)y) leq tf(x) + (1-t)f(y) - k = t(f(x)-k) + (1-t)(f(y)-k).$$
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
$endgroup$
add a comment |
$begingroup$
I think you just have an arithmetic error; namely,
$$s(tx + (1-t)y) leq tf(x) + (1-t)f(y) - k = t(f(x)-k) + (1-t)(f(y)-k).$$
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
$endgroup$
add a comment |
$begingroup$
I think you just have an arithmetic error; namely,
$$s(tx + (1-t)y) leq tf(x) + (1-t)f(y) - k = t(f(x)-k) + (1-t)(f(y)-k).$$
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
$endgroup$
I think you just have an arithmetic error; namely,
$$s(tx + (1-t)y) leq tf(x) + (1-t)f(y) - k = t(f(x)-k) + (1-t)(f(y)-k).$$
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
answered Mar 17 at 22:49
Gary MoonGary Moon
84616
84616
add a comment |
add a comment |
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