Is this function convex?
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From: ben on 28 May 2010 06:38 Let x1, x2,...xm be positive variables (actually, >= 1). The function is: 1/(x1*x2*...*xm). Intuitively it's gotta be convex, but I can't prove it. Tried every test for convexity I can find, including proof by induction. But they become intractable (at least for me). Any suggestions?
From: David C. Ullrich on 28 May 2010 07:56 On Fri, 28 May 2010 03:38:17 -0700 (PDT), ben <bwcrain(a)gmail.com> wrote: >Let x1, x2,...xm be positive variables (actually, >= 1). The function >is: > >1/(x1*x2*...*xm). Let's call that f(x). >Intuitively it's gotta be convex, but I can't prove it. Tried every >test for convexity I can find, including proof by induction. But they >become intractable (at least for me). Any suggestions? Fix a and b in R^m with a_j > 0 for all j and let g(t) = f(a + bt) for real t. You need to show that g is convex near the origin. If you differentiate you het g'(t) = g(t) (- sum (b_j/(a_j + b_j t))). Hence g''(t) = g'(t) (- sum (b_j/(a_j + b_j t))) + g(t) (sum(b_j/(a_j + b_j t))^2) = g(t) (- sum (b_j/(a_j + b_j t)))^2 + g(t) (sum(b_j/(a_j + b_j t))^2); hence g''(t) >= 0 for small t.
From: Ray Vickson on 28 May 2010 12:57 On May 28, 3:38 am, ben <bwcr...(a)gmail.com> wrote: > Let x1, x2,...xm be positive variables (actually, >= 1). The function > is: > > 1/(x1*x2*...*xm). > > Intuitively it's gotta be convex, but I can't prove it. Tried every > test for convexity I can find, including proof by induction. But they > become intractable (at least for me). Any suggestions? The Hessian H of f = 1/(x1*x2*...*xm) has ith diagonal element = 2*yi^2 and (i,j) off-diagonal element = yi*yj, where yk = 1/xk. Thus, for column vector z = col[z1,z2,...,zm] we have the ith component of Hz as (Hz)_i = yi^2*zi + yi*<y,z>, where <y,z> = sum yi*zi is the inner product. Therefore, the quadratic form z^THz = sum (yi*zi)^2 + <y,z>^2 is > 0 if z is not the zero vector. Thus, f is strictly convex. R.G. Vickson
From: David C. Ullrich on 28 May 2010 13:16 On Fri, 28 May 2010 06:56:16 -0500, David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >On Fri, 28 May 2010 03:38:17 -0700 (PDT), ben <bwcrain(a)gmail.com> >wrote: > >>Let x1, x2,...xm be positive variables (actually, >= 1). The function >>is: >> >>1/(x1*x2*...*xm). > >Let's call that f(x). > >>Intuitively it's gotta be convex, but I can't prove it. Tried every >>test for convexity I can find, including proof by induction. But they >>become intractable (at least for me). Any suggestions? Glad I gpt back here before someone else pointed out that there's a much better solution then the one below: It's clear from the definition "f(convex combination) <= convex combination(f)" that if L is convex and phi is increasing and convex then phi o L is convex. Hence it's clear that if f > 0 and log(f) is convex then f is convex... >Fix a and b in R^m with a_j > 0 for all j and let > > g(t) = f(a + bt) > >for real t. You need to show that g is convex near the origin. > >If you differentiate you het > > g'(t) = g(t) (- sum (b_j/(a_j + b_j t))). > >Hence > > g''(t) = g'(t) (- sum (b_j/(a_j + b_j t))) > + g(t) (sum(b_j/(a_j + b_j t))^2) > > = g(t) (- sum (b_j/(a_j + b_j t)))^2 > + g(t) (sum(b_j/(a_j + b_j t))^2); > >hence g''(t) >= 0 for small t. > > > > >
From: Ray Vickson on 28 May 2010 13:22
On May 28, 9:57 am, Ray Vickson <RGVick...(a)shaw.ca> wrote: > On May 28, 3:38 am, ben <bwcr...(a)gmail.com> wrote: > > > Let x1, x2,...xm be positive variables (actually, >= 1). The function > > is: > > > 1/(x1*x2*...*xm). > > > Intuitively it's gotta be convex, but I can't prove it. Tried every > > test for convexity I can find, including proof by induction. But they > > become intractable (at least for me). Any suggestions? > > The Hessian H of f = 1/(x1*x2*...*xm) has ith diagonal element = > 2*yi^2 and (i,j) off-diagonal element = yi*yj, where yk = 1/xk. Sorry: that is a typo. The matrix J = (x1*x2*...*xm)*H is the one with the elements specified above. Thus, z^T H z = z^T J z/(x1*x2*...*xm) = sum(yi*zi)^2 + <y,z>^2]/(x1*x2*...*xm). This is > 0 for all nonzero z- vectors as long as x is in the positive orthant. R.G. Vickson >Thus, > for column vector z = col[z1,z2,...,zm] we have the ith component of > Hz as (Hz)_i = yi^2*zi + yi*<y,z>, where <y,z> = sum yi*zi is the > inner product. Therefore, the quadratic form z^THz = sum (yi*zi)^2 + > <y,z>^2 is > 0 if z is not the zero vector. Thus, f is strictly > convex. > > R.G. Vickson |