Equivalence of convex function
$begingroup$
Let $I subseteq mathbb{R}$ be an interval. Then a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ when:
$$forall a,b in I, a < x < b Rightarrow f(x) le f(a) + dfrac{f(b)-f(a)}{b-a}.(x-a)$$
Also I know that a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ iff given $a,b in I, a < b$, the following happens:
$$f(t.a + (1-t).b) le t.f(a) + (1-t).f(b) forall t in [0,1]$$
There is another equivalence: A function $f$ is convex on the interval $I subseteq mathbb{R}$ iff given $a_1,a_2,...,a_n in I$ and $t_1,t_2,...,t_n in [0,1]$ that $sum t_i = 1$ we have:
$$f(sum t_i a_i) le sum t_i f(a_i)$$
How to prove this last equivalence?
real-analysis functions convex-analysis
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
$endgroup$
add a comment |
$begingroup$
Let $I subseteq mathbb{R}$ be an interval. Then a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ when:
$$forall a,b in I, a < x < b Rightarrow f(x) le f(a) + dfrac{f(b)-f(a)}{b-a}.(x-a)$$
Also I know that a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ iff given $a,b in I, a < b$, the following happens:
$$f(t.a + (1-t).b) le t.f(a) + (1-t).f(b) forall t in [0,1]$$
There is another equivalence: A function $f$ is convex on the interval $I subseteq mathbb{R}$ iff given $a_1,a_2,...,a_n in I$ and $t_1,t_2,...,t_n in [0,1]$ that $sum t_i = 1$ we have:
$$f(sum t_i a_i) le sum t_i f(a_i)$$
How to prove this last equivalence?
real-analysis functions convex-analysis
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
$endgroup$
1
$begingroup$
Induction. ${}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 3:09
$begingroup$
@T.Bongers Can you give further hints? I still don't know how to prove it.
$endgroup$
– Enzo Nakamura
Dec 4 '18 at 18:00
add a comment |
$begingroup$
Let $I subseteq mathbb{R}$ be an interval. Then a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ when:
$$forall a,b in I, a < x < b Rightarrow f(x) le f(a) + dfrac{f(b)-f(a)}{b-a}.(x-a)$$
Also I know that a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ iff given $a,b in I, a < b$, the following happens:
$$f(t.a + (1-t).b) le t.f(a) + (1-t).f(b) forall t in [0,1]$$
There is another equivalence: A function $f$ is convex on the interval $I subseteq mathbb{R}$ iff given $a_1,a_2,...,a_n in I$ and $t_1,t_2,...,t_n in [0,1]$ that $sum t_i = 1$ we have:
$$f(sum t_i a_i) le sum t_i f(a_i)$$
How to prove this last equivalence?
real-analysis functions convex-analysis
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
$endgroup$
Let $I subseteq mathbb{R}$ be an interval. Then a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ when:
$$forall a,b in I, a < x < b Rightarrow f(x) le f(a) + dfrac{f(b)-f(a)}{b-a}.(x-a)$$
Also I know that a function $f:I rightarrow mathbb{R}$ is convex on the interval $I subseteq mathbb{R}$ iff given $a,b in I, a < b$, the following happens:
$$f(t.a + (1-t).b) le t.f(a) + (1-t).f(b) forall t in [0,1]$$
There is another equivalence: A function $f$ is convex on the interval $I subseteq mathbb{R}$ iff given $a_1,a_2,...,a_n in I$ and $t_1,t_2,...,t_n in [0,1]$ that $sum t_i = 1$ we have:
$$f(sum t_i a_i) le sum t_i f(a_i)$$
How to prove this last equivalence?
real-analysis functions convex-analysis
real-analysis functions convex-analysis
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
edited Dec 4 '18 at 5:27
GNUSupporter 8964民主女神 地下教會
13.5k72550
13.5k72550
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
asked Dec 4 '18 at 3:09
Enzo NakamuraEnzo Nakamura
685
685
1
$begingroup$
Induction. ${}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 3:09
$begingroup$
@T.Bongers Can you give further hints? I still don't know how to prove it.
$endgroup$
– Enzo Nakamura
Dec 4 '18 at 18:00
add a comment |
1
$begingroup$
Induction. ${}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 3:09
$begingroup$
@T.Bongers Can you give further hints? I still don't know how to prove it.
$endgroup$
– Enzo Nakamura
Dec 4 '18 at 18:00
1
1
$begingroup$
Induction. ${}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 3:09
$begingroup$
Induction. ${}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 3:09
$begingroup$
@T.Bongers Can you give further hints? I still don't know how to prove it.
$endgroup$
– Enzo Nakamura
Dec 4 '18 at 18:00
$begingroup$
@T.Bongers Can you give further hints? I still don't know how to prove it.
$endgroup$
– Enzo Nakamura
Dec 4 '18 at 18:00
add a comment |
0
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$begingroup$
Induction. ${}{}$
$endgroup$
– T. Bongers
Dec 4 '18 at 3:09
$begingroup$
@T.Bongers Can you give further hints? I still don't know how to prove it.
$endgroup$
– Enzo Nakamura
Dec 4 '18 at 18:00