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Let $y,\lambda\in\mathbb{R}^n$. I want to minimize the following with respect to $y$. $$ f(y)=||y|| + \lambda^Ty $$ where $||y||$ is the Euclidean norm. I first take the derivative of the function and get $$ \nabla f(y) = \frac{y}{||y||} + \lambda $$ Then I tried to get the Hessian matrix but I could not conclude anything. Is $f(y)$ is a convex function? If not how can I find the minimum of this function?

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  • $\begingroup$ It is convex, because it is the sum of two convex functions. $\endgroup$ – Tomás Feb 2 '13 at 16:16
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    $\begingroup$ Baby case: $n=1$. The minimum is $0$ if $-1\leq \lambda\leq 1$, and there is no minimum otherwise as the infimum is $-\infty$. $\endgroup$ – Julien Feb 2 '13 at 16:19
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Note that $$\|y\|+\langle \lambda,y\rangle\geq \|y\|(1-\|\lambda\|) $$

From the last inequality we conclude that if $\|\lambda\|\in [0,1]$, then the minimum of $f$ is zero.

On the other hand, if $\|\lambda\|>1$, you can take $y=-t\lambda$, and make $t\rightarrow\infty$ to conclude that $f(-t\lambda)\rightarrow-\infty$.

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  • $\begingroup$ From the last inequality if $||\lambda|| \in [0,1]$, then the minimum of the function in the right is zero. How can you conclude that $f$ is also zero? $\endgroup$ – neticin Feb 2 '13 at 16:32
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    $\begingroup$ Note that $f(0)=0$. $\endgroup$ – Julien Feb 2 '13 at 16:35
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If $y'$ is a minimum and is not 0, then let $c>0$,

Then $f(cy') = ||cy'|| + c\lambda^Ty'=c( ||y'|| + \lambda^Ty') $

This function is zero at zero, so you're not gonna get a minimum that is less than zero by this calculation.

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