Proving that a strongly convex function is coercive












5












$begingroup$


I am having trouble with this proof. I am given the following 2 definitions:




1) A function $f$ is coercive if $lim_{||x|| rightarrow infty} f(x) = infty$



2) A $C^2$ function $f$ is strongly convex if there exists a constant $c_0 > 0$ such that: $(x - y)^T (nabla f(x) - nabla f(y)) geq c_0 ||x - y||^2 hspace{5mm}$ $forall x,y in mathbb{R}^n$




The question is to show that if $f$ is strongly convex then it is coercive.



I am only allowed to use basic theorems such as Taylor expansion, triangle inequality etc. However, I can use the fact that a function is coercive $iff$ all its level sets are compact.



My instinct is that the answer can be obtained by a doing a Taylor expansion and manipulating the result, but I've been stuck for days using this approach. Any help would be greatly appreciated.






multivariable-calculus convex-optimization nonlinear-optimization coercive







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  • $begingroup$
    Is there anything wrong with my answer ?
    $endgroup$
    – Gabriel Romon
    Mar 22 '17 at 14:30
















5












$begingroup$


I am having trouble with this proof. I am given the following 2 definitions:




1) A function $f$ is coercive if $lim_{||x|| rightarrow infty} f(x) = infty$



2) A $C^2$ function $f$ is strongly convex if there exists a constant $c_0 > 0$ such that: $(x - y)^T (nabla f(x) - nabla f(y)) geq c_0 ||x - y||^2 hspace{5mm}$ $forall x,y in mathbb{R}^n$




The question is to show that if $f$ is strongly convex then it is coercive.



I am only allowed to use basic theorems such as Taylor expansion, triangle inequality etc. However, I can use the fact that a function is coercive $iff$ all its level sets are compact.



My instinct is that the answer can be obtained by a doing a Taylor expansion and manipulating the result, but I've been stuck for days using this approach. Any help would be greatly appreciated.






multivariable-calculus convex-optimization nonlinear-optimization coercive







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there anything wrong with my answer ?
    $endgroup$
    – Gabriel Romon
    Mar 22 '17 at 14:30














5












5








5


4



$begingroup$


I am having trouble with this proof. I am given the following 2 definitions:




1) A function $f$ is coercive if $lim_{||x|| rightarrow infty} f(x) = infty$



2) A $C^2$ function $f$ is strongly convex if there exists a constant $c_0 > 0$ such that: $(x - y)^T (nabla f(x) - nabla f(y)) geq c_0 ||x - y||^2 hspace{5mm}$ $forall x,y in mathbb{R}^n$




The question is to show that if $f$ is strongly convex then it is coercive.



I am only allowed to use basic theorems such as Taylor expansion, triangle inequality etc. However, I can use the fact that a function is coercive $iff$ all its level sets are compact.



My instinct is that the answer can be obtained by a doing a Taylor expansion and manipulating the result, but I've been stuck for days using this approach. Any help would be greatly appreciated.






multivariable-calculus convex-optimization nonlinear-optimization coercive







share|cite|improve this question











$endgroup$




I am having trouble with this proof. I am given the following 2 definitions:




1) A function $f$ is coercive if $lim_{||x|| rightarrow infty} f(x) = infty$



2) A $C^2$ function $f$ is strongly convex if there exists a constant $c_0 > 0$ such that: $(x - y)^T (nabla f(x) - nabla f(y)) geq c_0 ||x - y||^2 hspace{5mm}$ $forall x,y in mathbb{R}^n$




The question is to show that if $f$ is strongly convex then it is coercive.



I am only allowed to use basic theorems such as Taylor expansion, triangle inequality etc. However, I can use the fact that a function is coercive $iff$ all its level sets are compact.



My instinct is that the answer can be obtained by a doing a Taylor expansion and manipulating the result, but I've been stuck for days using this approach. Any help would be greatly appreciated.






multivariable-calculus convex-optimization nonlinear-optimization coercive




multivariable-calculus convex-optimization nonlinear-optimization coercive






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share|cite|improve this question













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edited Mar 19 '17 at 9:45







r.emp

















asked Mar 18 '17 at 20:28









r.empr.emp

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  • $begingroup$
    Is there anything wrong with my answer ?
    $endgroup$
    – Gabriel Romon
    Mar 22 '17 at 14:30


















  • $begingroup$
    Is there anything wrong with my answer ?
    $endgroup$
    – Gabriel Romon
    Mar 22 '17 at 14:30
















$begingroup$
Is there anything wrong with my answer ?
$endgroup$
– Gabriel Romon
Mar 22 '17 at 14:30




$begingroup$
Is there anything wrong with my answer ?
$endgroup$
– Gabriel Romon
Mar 22 '17 at 14:30










1 Answer
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$begingroup$

Lemma 1: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, (nabla f(y)-nabla f(x))^T(y-x)geq 0$.



Lemma 2: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, f(y)geq f(x)+nabla f(x)^T(y-x)$




Let us prove first that your definition of "strongly convex" implies convexity of $displaystyle g:xmapsto f(x)-frac{c_0}2|x|^2$



Indeed, $begin{aligned}[t](nabla g(y)-nabla g(x))^T(y-x)&=(nabla f(y)-c_0y - (nabla f(x)-c_0x))^T(y-x)\
&=(nabla f(y)-nabla f(x))^T(y-x) - (c_0y-c_0x)^T(y-x) \
&= (nabla f(y)-nabla f(x))^T(y-x)-c_0|y-x|^2 \
&geq 0 quad text{ with your definition}
end{aligned}$



By Lemma 1, $g$ is convex.




By Lemma 2, $$begin{aligned}[t]
g(y)&geq g(x)+nabla g(x)^T(y-x) \
f(y)-frac{c_0}2|y|^2 &geq f(x)-frac{c_0}2|x|^2 + (nabla f(x)-c_0x)^T(y-x)
end{aligned}$$
Let $alpha =nabla f(x)-c_0x$. Then
$$f(y)geq frac{c_0}2|y|^2+ alpha^Ty+K_x$$ where $K_x$ is a constant that depends only on $x$.



By Cauchy-Schwarz, $alpha^Ty geq -|alpha||y|$, hence
$$f(y)geq frac{c_0}2|y|^2 -|alpha||y|+K_x$$



The right-hand side goes to $infty$ as $|y|to infty$ and we're done.






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    $begingroup$

    Lemma 1: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, (nabla f(y)-nabla f(x))^T(y-x)geq 0$.



    Lemma 2: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, f(y)geq f(x)+nabla f(x)^T(y-x)$




    Let us prove first that your definition of "strongly convex" implies convexity of $displaystyle g:xmapsto f(x)-frac{c_0}2|x|^2$



    Indeed, $begin{aligned}[t](nabla g(y)-nabla g(x))^T(y-x)&=(nabla f(y)-c_0y - (nabla f(x)-c_0x))^T(y-x)\
    &=(nabla f(y)-nabla f(x))^T(y-x) - (c_0y-c_0x)^T(y-x) \
    &= (nabla f(y)-nabla f(x))^T(y-x)-c_0|y-x|^2 \
    &geq 0 quad text{ with your definition}
    end{aligned}$



    By Lemma 1, $g$ is convex.




    By Lemma 2, $$begin{aligned}[t]
    g(y)&geq g(x)+nabla g(x)^T(y-x) \
    f(y)-frac{c_0}2|y|^2 &geq f(x)-frac{c_0}2|x|^2 + (nabla f(x)-c_0x)^T(y-x)
    end{aligned}$$
    Let $alpha =nabla f(x)-c_0x$. Then
    $$f(y)geq frac{c_0}2|y|^2+ alpha^Ty+K_x$$ where $K_x$ is a constant that depends only on $x$.



    By Cauchy-Schwarz, $alpha^Ty geq -|alpha||y|$, hence
    $$f(y)geq frac{c_0}2|y|^2 -|alpha||y|+K_x$$



    The right-hand side goes to $infty$ as $|y|to infty$ and we're done.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Lemma 1: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, (nabla f(y)-nabla f(x))^T(y-x)geq 0$.



      Lemma 2: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, f(y)geq f(x)+nabla f(x)^T(y-x)$




      Let us prove first that your definition of "strongly convex" implies convexity of $displaystyle g:xmapsto f(x)-frac{c_0}2|x|^2$



      Indeed, $begin{aligned}[t](nabla g(y)-nabla g(x))^T(y-x)&=(nabla f(y)-c_0y - (nabla f(x)-c_0x))^T(y-x)\
      &=(nabla f(y)-nabla f(x))^T(y-x) - (c_0y-c_0x)^T(y-x) \
      &= (nabla f(y)-nabla f(x))^T(y-x)-c_0|y-x|^2 \
      &geq 0 quad text{ with your definition}
      end{aligned}$



      By Lemma 1, $g$ is convex.




      By Lemma 2, $$begin{aligned}[t]
      g(y)&geq g(x)+nabla g(x)^T(y-x) \
      f(y)-frac{c_0}2|y|^2 &geq f(x)-frac{c_0}2|x|^2 + (nabla f(x)-c_0x)^T(y-x)
      end{aligned}$$
      Let $alpha =nabla f(x)-c_0x$. Then
      $$f(y)geq frac{c_0}2|y|^2+ alpha^Ty+K_x$$ where $K_x$ is a constant that depends only on $x$.



      By Cauchy-Schwarz, $alpha^Ty geq -|alpha||y|$, hence
      $$f(y)geq frac{c_0}2|y|^2 -|alpha||y|+K_x$$



      The right-hand side goes to $infty$ as $|y|to infty$ and we're done.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Lemma 1: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, (nabla f(y)-nabla f(x))^T(y-x)geq 0$.



        Lemma 2: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, f(y)geq f(x)+nabla f(x)^T(y-x)$




        Let us prove first that your definition of "strongly convex" implies convexity of $displaystyle g:xmapsto f(x)-frac{c_0}2|x|^2$



        Indeed, $begin{aligned}[t](nabla g(y)-nabla g(x))^T(y-x)&=(nabla f(y)-c_0y - (nabla f(x)-c_0x))^T(y-x)\
        &=(nabla f(y)-nabla f(x))^T(y-x) - (c_0y-c_0x)^T(y-x) \
        &= (nabla f(y)-nabla f(x))^T(y-x)-c_0|y-x|^2 \
        &geq 0 quad text{ with your definition}
        end{aligned}$



        By Lemma 1, $g$ is convex.




        By Lemma 2, $$begin{aligned}[t]
        g(y)&geq g(x)+nabla g(x)^T(y-x) \
        f(y)-frac{c_0}2|y|^2 &geq f(x)-frac{c_0}2|x|^2 + (nabla f(x)-c_0x)^T(y-x)
        end{aligned}$$
        Let $alpha =nabla f(x)-c_0x$. Then
        $$f(y)geq frac{c_0}2|y|^2+ alpha^Ty+K_x$$ where $K_x$ is a constant that depends only on $x$.



        By Cauchy-Schwarz, $alpha^Ty geq -|alpha||y|$, hence
        $$f(y)geq frac{c_0}2|y|^2 -|alpha||y|+K_x$$



        The right-hand side goes to $infty$ as $|y|to infty$ and we're done.






        share|cite|improve this answer









        $endgroup$



        Lemma 1: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, (nabla f(y)-nabla f(x))^T(y-x)geq 0$.



        Lemma 2: Let $f:mathbb R^nto mathbb R$ be a Fréchet-differentiable function. $f$ is convex iff $forall x,yin mathbb R^n, f(y)geq f(x)+nabla f(x)^T(y-x)$




        Let us prove first that your definition of "strongly convex" implies convexity of $displaystyle g:xmapsto f(x)-frac{c_0}2|x|^2$



        Indeed, $begin{aligned}[t](nabla g(y)-nabla g(x))^T(y-x)&=(nabla f(y)-c_0y - (nabla f(x)-c_0x))^T(y-x)\
        &=(nabla f(y)-nabla f(x))^T(y-x) - (c_0y-c_0x)^T(y-x) \
        &= (nabla f(y)-nabla f(x))^T(y-x)-c_0|y-x|^2 \
        &geq 0 quad text{ with your definition}
        end{aligned}$



        By Lemma 1, $g$ is convex.




        By Lemma 2, $$begin{aligned}[t]
        g(y)&geq g(x)+nabla g(x)^T(y-x) \
        f(y)-frac{c_0}2|y|^2 &geq f(x)-frac{c_0}2|x|^2 + (nabla f(x)-c_0x)^T(y-x)
        end{aligned}$$
        Let $alpha =nabla f(x)-c_0x$. Then
        $$f(y)geq frac{c_0}2|y|^2+ alpha^Ty+K_x$$ where $K_x$ is a constant that depends only on $x$.



        By Cauchy-Schwarz, $alpha^Ty geq -|alpha||y|$, hence
        $$f(y)geq frac{c_0}2|y|^2 -|alpha||y|+K_x$$



        The right-hand side goes to $infty$ as $|y|to infty$ and we're done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 19 '17 at 11:09









        Gabriel RomonGabriel Romon

        17.9k53284




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