MHB Show that a linear function is convex

[mathmari]

Hey!
To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?

 

[I like Serena]

Re: show that a linear function is convex

Hey!
To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?

Hi!

Is the Hessian matrix positive semi-definite?
Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?

 

[mathmari]

Re: show that a linear function is convex

Hi!

Is the Hessian matrix positive semi-definite?
Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?

for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $ H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}] $. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite. But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?

 

[I like Serena]

Re: show that a linear function is convex

for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $ H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}] $. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite.

Yep.
(Although you should leave out the absolute value symbols for $D1$. )

But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?

Positive definite requires $>0$, which is not the case.
Similarly negative definite requires $<0$, which is also not the case.

So if the hessian matrix is the zero matrix it is neither positive definite nor negative definite.
However, it is both positive semi-definite and negative semi-definite.

 

[mathmari]

Re: show that a linear function is convex

However, it is both positive semi-definite and negative semi-definite.

so do we conlude that the function is both concave and convex??

 

[I like Serena]

Re: show that a linear function is convex

so do we conlude that the function is both concave and convex??

Yes.
Note that it is neither strictly convex, nor strictly concave.

 

[mathmari]

Re: show that a linear function is convex

Yes.
Note that it is neither strictly convex, nor strictly concave.

Ok! Thank you!

 

[Deveno]

I believe the technical term here is "flat" (;)) (although "hyper-planar" has a nicer ring to it, n'est-ce pas?).

 

[mathmari]

I believe the technical term here is "flat" (;)) (although "hyper-planar" has a nicer ring to it, n'est-ce pas?).

Do you mean that this is the technical term that a function is both concave and convex?