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Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently(sufficiently small) fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)?

My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or perhaps there is some counter-example to this claim?

Thanks in advance!

Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently small fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)?

My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or perhaps there is some counter-example to this claim?

Thanks in advance!

Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a (sufficiently small) fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)?

My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or perhaps there is some counter-example to this claim?

Thanks in advance!

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Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently small fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? 

My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or is perhaps there is some counter-example to this claim?

Thanks in advance!

Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently small fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or is perhaps there some counter-example to this claim?

Thanks in advance!

Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently small fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? 

My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or perhaps there is some counter-example to this claim?

Thanks in advance!

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Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a smooth and function which is convex functionand smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently small fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or is perhaps there some counter-example to this claim?

Thanks in advance!

Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a smooth and convex function. If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently small fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or is perhaps there some counter-example to this claim?

Thanks in advance!

Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a sufficiently small fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$.

Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or is perhaps there some counter-example to this claim?

Thanks in advance!

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