Convexity of a functional on a Sobolev space

Question

Denote a specific Sobolev space by $$W^{2,2}(a,b) = \left\lbrace x\in L_2(a,b) : x'\in AC[a,b],\quad x''\in L_2(a,b) \right\rbrace $$ where $AC[a,b]$ is the class of absolutely continuous functions on $[a,b]\subset\mathbb R$.

Given smoothing problems in the context of theory of splines (and, as I've witnessed, in statistics) the objective is to minimize the following expression over all $f\in W^{2,2}$: $$M(f) := \left [ \sum_{j=0}^n w_j(f(x_j)-y_j)^2 + \int_a^bf''(x)^2w(x)dx\right ]^{1/2} $$ where $x_0<x_1<\ldots <x_n$ (in $[a,b]$), $w,w_j>0$ (weights) and $y_j\in\mathbb R$ are provided and then the (unique?) minimizer of $M$ is christened the smoothing spline.

What about the functional $M : W^{2,2}\to\mathbb R$ itself, though? Is it, perhaps, convex? Even strictly convex?

Specifically, for convexity: $$\forall u,v\in W^{2,2}, \forall t\in (0,1), M(tu +(1-t)v)\overset{?}\leq tM(u)+(1-t)M(v).$$ Strict convexity means that whenever $u\neq v$ the inequality is strict.

I haven't been able to find any discussion on this particular property of $M$ and it doesn't seem all that obvious either.

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A thought. Perhaps, we could obtain $$M^2(tu+(1-t)v)\leq t^2M^2(u)+(1-t)^2M^2(v) $$ and state that perhaps $2t(1-t)M(u)M(v)>0$ for $u\neq v$.

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That won't work. If $u = 0$ with $y_j\equiv 0$, then $M(u) = 0$ and at the same time $$M((1-t)v) = (1-t)^2M(v)< (1-t)M(v), $$ since $0<t<1$, provided $M(v)>0$.

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Answer 1

Here is how I use Minkowski inequality in both discrete and continuous setting

Let $$A(f) := \left [ \sum_{j=0}^n w_j(f(x_j)-y_j)^2 \right ]^{1/2}~~~~and ~~~~B(f) := \left [ \int_a^bf''(x)^2 w(x)dx\right ]^{1/2}$$

Then $$M(f) := \left [ \sum_{j=0}^n w_j(f(x_j)-y_j)^2 + \int_a^bf''(x)^2w(x)dx\right ]^{1/2} =\left [ A^2(f) +B^2(f)\right ]^{1/2}$$

Let $0<t<1$ and $g\neq f$ then,

• Looking at the Hilbert spaces $$\ell^2(w) = \left\{z=(z_i)_i: \|z\|^2_{\ell^2(w)} =\sum_{i=0}^{n} w_i |z_i|^2<\infty\right\}$$

we set $z^t = (z^t_i)_{0\le i\le n}$ and $\xi^t = (\xi^t_i)_{0\le i\le n}$ with $$z^t_i=t(f(x_j)-y_j) ~~~and~~~~\xi^t_i=(1-t)(g(x_j)-y_j)$$

By Minkowski inequality we have $$A(tf+(1-t)g) := \left [ \sum_{j=0}^n w_j(t(f(x_j)-y_j) +(1-t)(g(x_j)-y_j))^2 \right ]^{1/2}\\= \|z^t+\xi^t\|_{\ell^2(w)}\\< \|z^t\|_{\ell^2(w)}+\|\xi^t\|_{\ell^2(w)} \\= tA(f)+(1-t)A(g) $$

• Similarly, using the again the Minkowski inequality for integrals we have,

$$B(tf+(1-t)g) = \int_a^b \left[tf''(x) +(1-t)g''(x)\right]^2w(x)dx \le tB(f)+(1-t)B(g)$$

Finally, applying Minkowski inequality in $\color{blue}{\Bbb R^2}$ for vectors $(tA(f),tB(f))^T$ and $((1-t)A(f),(1-t)B(f))^T$ together with the foregoing, we get,

\begin{align}\color{red}{M(tf+(1-t)g) }&= \left[A^2(tf+(1-t)g)+B^2(tf+(1-t)g)\right]^{1/2}\\&\le \left[(tA(f)+(1-t)A(g))^2+(tB(f)+(1-t)B(g))^2\right]^{1/2}\\&\le \left[(tA(f)+tB(g))^2+((1-t)B(f)+(1-t)B(g))^2\right]^{1/2} \\&=\color{red}{ tM(f)+(1-t)M(g)}.\end{align} this prove the convexity of $M$.

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