Mixing
Prove derivation of global minimum of a convex function is zero (using only convexity)
$begingroup$
I am reading the Wikipedia of Convex function. in it, it is written:
if $f$ is a convex function then $$ f(x) ge f(y) + f'(y)(x-y)$$
for all $x$ and $y$ in the interval. In particular, if $f ′(c) = 0$, then $c$
is a global minimum of $f(x)$.
I fully understand this, but I'm wondering if it is possible to prove that if $c$ if the point of the minimum then $f ′(c) = 0$ just by using convexity. I think it is possible but until now I could not prove that.
derivatives optimization convex-analysis
edited Jan 1 at 21:15
A.Γ.
22.6k32656
asked Jan 1 at 21:13
PeymanPeyman
939
$endgroup$
|
show 5 more comments
$begingroup$
I am reading the Wikipedia of Convex function. in it, it is written:
if $f$ is a convex function then $$ f(x) ge f(y) + f'(y)(x-y)$$
for all $x$ and $y$ in the interval. In particular, if $f ′(c) = 0$, then $c$
is a global minimum of $f(x)$.
I fully understand this, but I'm wondering if it is possible to prove that if $c$ if the point of the minimum then $f ′(c) = 0$ just by using convexity. I think it is possible but until now I could not prove that.
derivatives optimization convex-analysis
edited Jan 1 at 21:15
A.Γ.
22.6k32656
asked Jan 1 at 21:13
PeymanPeyman
939
$endgroup$
1
$begingroup$
If $c$ is a (local) minimum then $f'(c)=0$ for any differentiable function without convexity. Do you mean to prove the other way round?
$endgroup$
– A.Γ.
Jan 1 at 21:17
$begingroup$
@A.Γ. yes I mean to prove that just by the definition of convex functions.
$endgroup$
– Peyman
Jan 1 at 21:19
$begingroup$
Is your question Fermat's theorem?
$endgroup$
– A.Γ.
Jan 1 at 22:23
$begingroup$
@A.Γ. not it is not. see, I know that for any function If c is a (local) minimum then $f′(c)=0$. but in this case, I want to prove that for convex functions just by using its definition, and I don't know why I can't. I think there must be a way to prove that easily for both functions of one variable and functions of $n$ variables.
$endgroup$
– Peyman
Jan 1 at 22:36
1
$begingroup$
If $f(x) ge f(c)$ for all $x$ then the above inequality shows that $f'(c) (x-c) le 0$ for all $x$ and hence $f'(c) = 0$.
$endgroup$
– copper.hat
Jan 1 at 23:02
|
show 5 more comments
$begingroup$
I am reading the Wikipedia of Convex function. in it, it is written:
if $f$ is a convex function then $$ f(x) ge f(y) + f'(y)(x-y)$$
for all $x$ and $y$ in the interval. In particular, if $f ′(c) = 0$, then $c$
is a global minimum of $f(x)$.
I fully understand this, but I'm wondering if it is possible to prove that if $c$ if the point of the minimum then $f ′(c) = 0$ just by using convexity. I think it is possible but until now I could not prove that.
derivatives optimization convex-analysis
edited Jan 1 at 21:15
A.Γ.
22.6k32656
asked Jan 1 at 21:13
PeymanPeyman
939
$endgroup$
I am reading the Wikipedia of Convex function. in it, it is written:
if $f$ is a convex function then $$ f(x) ge f(y) + f'(y)(x-y)$$
for all $x$ and $y$ in the interval. In particular, if $f ′(c) = 0$, then $c$
is a global minimum of $f(x)$.
I fully understand this, but I'm wondering if it is possible to prove that if $c$ if the point of the minimum then $f ′(c) = 0$ just by using convexity. I think it is possible but until now I could not prove that.
derivatives optimization convex-analysis
derivatives optimization convex-analysis
edited Jan 1 at 21:15
A.Γ.
22.6k32656
asked Jan 1 at 21:13
PeymanPeyman
939
edited Jan 1 at 21:15
A.Γ.
22.6k32656
asked Jan 1 at 21:13
PeymanPeyman
939
edited Jan 1 at 21:15
A.Γ.
22.6k32656
edited Jan 1 at 21:15
A.Γ.
22.6k32656
edited Jan 1 at 21:15
A.Γ.
22.6k32656
22.6k32656
asked Jan 1 at 21:13
PeymanPeyman
939
asked Jan 1 at 21:13
PeymanPeyman
939
asked Jan 1 at 21:13
PeymanPeyman
939
939
1
$begingroup$
If $c$ is a (local) minimum then $f'(c)=0$ for any differentiable function without convexity. Do you mean to prove the other way round?
$endgroup$
– A.Γ.
Jan 1 at 21:17
$begingroup$
@A.Γ. yes I mean to prove that just by the definition of convex functions.
$endgroup$
– Peyman
Jan 1 at 21:19
$begingroup$
Is your question Fermat's theorem?
$endgroup$
– A.Γ.
Jan 1 at 22:23
$begingroup$
@A.Γ. not it is not. see, I know that for any function If c is a (local) minimum then $f′(c)=0$. but in this case, I want to prove that for convex functions just by using its definition, and I don't know why I can't. I think there must be a way to prove that easily for both functions of one variable and functions of $n$ variables.
$endgroup$
– Peyman
Jan 1 at 22:36
1
$begingroup$
If $f(x) ge f(c)$ for all $x$ then the above inequality shows that $f'(c) (x-c) le 0$ for all $x$ and hence $f'(c) = 0$.
$endgroup$
– copper.hat
Jan 1 at 23:02
|
show 5 more comments
1
$begingroup$
If $c$ is a (local) minimum then $f'(c)=0$ for any differentiable function without convexity. Do you mean to prove the other way round?
$endgroup$
– A.Γ.
Jan 1 at 21:17
$begingroup$
@A.Γ. yes I mean to prove that just by the definition of convex functions.
$endgroup$
– Peyman
Jan 1 at 21:19
$begingroup$
Is your question Fermat's theorem?
$endgroup$
– A.Γ.
Jan 1 at 22:23
$begingroup$
@A.Γ. not it is not. see, I know that for any function If c is a (local) minimum then $f′(c)=0$. but in this case, I want to prove that for convex functions just by using its definition, and I don't know why I can't. I think there must be a way to prove that easily for both functions of one variable and functions of $n$ variables.
$endgroup$
– Peyman
Jan 1 at 22:36
1
$begingroup$
If $f(x) ge f(c)$ for all $x$ then the above inequality shows that $f'(c) (x-c) le 0$ for all $x$ and hence $f'(c) = 0$.
$endgroup$
– copper.hat
Jan 1 at 23:02
1
1
$begingroup$
If $c$ is a (local) minimum then $f'(c)=0$ for any differentiable function without convexity. Do you mean to prove the other way round?
$endgroup$
– A.Γ.
Jan 1 at 21:17
$begingroup$
If $c$ is a (local) minimum then $f'(c)=0$ for any differentiable function without convexity. Do you mean to prove the other way round?
$endgroup$
– A.Γ.
Jan 1 at 21:17
$begingroup$
@A.Γ. yes I mean to prove that just by the definition of convex functions.
$endgroup$
– Peyman
Jan 1 at 21:19
$begingroup$
@A.Γ. yes I mean to prove that just by the definition of convex functions.
$endgroup$
– Peyman
Jan 1 at 21:19
$begingroup$
Is your question Fermat's theorem?
$endgroup$
– A.Γ.
Jan 1 at 22:23
$begingroup$
Is your question Fermat's theorem?
$endgroup$
– A.Γ.
Jan 1 at 22:23
$begingroup$
@A.Γ. not it is not. see, I know that for any function If c is a (local) minimum then $f′(c)=0$. but in this case, I want to prove that for convex functions just by using its definition, and I don't know why I can't. I think there must be a way to prove that easily for both functions of one variable and functions of $n$ variables.
$endgroup$
– Peyman
Jan 1 at 22:36
$begingroup$
@A.Γ. not it is not. see, I know that for any function If c is a (local) minimum then $f′(c)=0$. but in this case, I want to prove that for convex functions just by using its definition, and I don't know why I can't. I think there must be a way to prove that easily for both functions of one variable and functions of $n$ variables.
$endgroup$
– Peyman
Jan 1 at 22:36
1
1
$begingroup$
If $f(x) ge f(c)$ for all $x$ then the above inequality shows that $f'(c) (x-c) le 0$ for all $x$ and hence $f'(c) = 0$.
$endgroup$
– copper.hat
Jan 1 at 23:02
$begingroup$
If $f(x) ge f(c)$ for all $x$ then the above inequality shows that $f'(c) (x-c) le 0$ for all $x$ and hence $f'(c) = 0$.
$endgroup$
– copper.hat
Jan 1 at 23:02
|
show 5 more comments
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$begingroup$
If $c$ is a (local) minimum then $f'(c)=0$ for any differentiable function without convexity. Do you mean to prove the other way round?
$endgroup$
– A.Γ.
Jan 1 at 21:17
$begingroup$
@A.Γ. yes I mean to prove that just by the definition of convex functions.
$endgroup$
– Peyman
Jan 1 at 21:19
$begingroup$
Is your question Fermat's theorem?
$endgroup$
– A.Γ.
Jan 1 at 22:23
$begingroup$
@A.Γ. not it is not. see, I know that for any function If c is a (local) minimum then $f′(c)=0$. but in this case, I want to prove that for convex functions just by using its definition, and I don't know why I can't. I think there must be a way to prove that easily for both functions of one variable and functions of $n$ variables.
$endgroup$
– Peyman
Jan 1 at 22:36
1
$begingroup$
If $f(x) ge f(c)$ for all $x$ then the above inequality shows that $f'(c) (x-c) le 0$ for all $x$ and hence $f'(c) = 0$.
$endgroup$
– copper.hat
Jan 1 at 23:02