Minimizer of a multivariable function

Let $f:[0,\infty)^n\to \mathbb{R}$ be a function continuously differentiable in the interior $(0,\infty)^n$ and that $\frac{\partial}{\partial x_j}f(\textbf{x})\to -\infty$ as $x_j\to 0^+$ for $j=1,\dots,n$.

Can it be shown rigorously that when this function is minimized over a set determined by a linear equation say $\{\textbf{x}=(x_1,\dots,x_n):\sum_j a_j x_j=b, x_j\ge 0\}$, the minimizer doesn't have a $0$ entry at a position when the constraint set allows non zero entries for that position?

Thanks.

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The assumptions are contradictory: A continuously differentiable function would be differentiable at $x_j=0$, with $$\left[\frac{\partial}{\partial x_j}f(\textbf{x})\right]_{x_j=0}=\lim_{x_j\to0^+}\frac{\partial}{\partial x_j}f(\textbf{x})\;,$$ but this limit doesn't exist by assumption. – joriki Sep 20 '12 at 8:12
@joriki: Yes, you are right. I should have said "continuously differentiable in the interior and that $\lim_{x_j\to0^+}\frac{\partial}{\partial x_j}f(\textbf{x})=-\infty$". I will edit accordingly. Thank you. – Kumara Sep 20 '12 at 9:46

This is false. A counterexample is given by

$$f(x,y)=\sqrt[4]{(x-1)^2+y^2}-1.1\sqrt y$$

with the linear constraint $x+y=1$. The partial derivative with respect to $y$ is

$$\frac{\partial f}{\partial y} = \frac y{2\left((x-1)^2+y^2\right)^{3/4}}-\frac{1.1}{2\sqrt y}\;,$$

which goes to $-\infty$ as $y\to0$ for fixed $x$ (including $x=1$). On the line $x+y=1$, we have $x=1-y$ and

$$f(1-y,y)=\sqrt[4]{y^2+y^2}-1.1\sqrt y=\left(\sqrt[4]2-1.1\right)\sqrt y\;,$$

which is minimal for $y=0$.

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Yes your counterexample disproves my claim. Can you say something on what additional condition on $f$ will make my claim true? One thing I forgot to mention is that my $f$ is non negative and quasi convex. I don't know whether these are relevent. – Kumara Sep 20 '12 at 10:42
I doubt that these two are relevant -- I think my function is quasi-convex (at least some plots by Wolfram|Alpha suggest that it is), and it could easily be made non-negative without destroying the minimum at $y=0$ (though I don't know if this could be done while keeping it quasi-convex). I think the condition that the partial derivatives parallel to the boundary must be finite should suffice; my counterexample uses an infinite partial derivative with respect to $x$ to counter the infinite partial derivative with respect to $y$. – joriki Sep 20 '12 at 11:00
I couldn't understand your last statement ...I think the condition that the partial derivatives parallel to the boundary must be finite should suffice; my counterexample uses an infinite partial derivative with respect to x to counter the infinite partial derivative with respect to y...Could you please elaborate? – Kumara Sep 20 '12 at 11:07
@Kumara: In my counterexample, the partial derivative with respect to $x$ diverges at $(1,0)$. This is what allows the function value to increase along the line $x+y=1$ even though it decreases infinitely fast along the line $x=1$. If the partial derivative with respect to $x$ at that point where finite, the infinite decrease with respect to $y$ would "win". – joriki Sep 20 '12 at 11:10
I am still not very clear about this. Could you please state the sufficient conditions for the minimizer to be in the relative interior of the set? – Kumara Oct 19 '12 at 9:40