Let $\mathbb{X}$ be a Banach space equipped with some norm $||\cdot||_\mathbb{X}$ and $F:\mathbb{X}\to\mathbb{R}$ be some linear functional. Suppose we are given a set $A\subseteq\mathbb{X}$ which is convex and uniformly bounded i.e. $\sup_{f\in A}||f||_{\mathbb{X}}<+\infty$. Consider the following optimization problem $$\inf_{f\in A} F(f):=\mu$$ My question is: If we know that the infimum of $F$ on $A$ is finite but not attainable, how do we estimate this infimum value? I know that in such a case it is desirable to construct a sequence $(f_k)\subseteq A$ such that $\lim_kF(f_k)=\mu$ i.e. construct a minimizing sequence. If we are lucky enough that $F(f_k)$ are easily computable then we just look at the limit of a sequence of real numbers $(F(f_k))$. However are there explicit iterative methods for constructing such sequences? What relevant research is done in this direction?

Some context: This question is related to this specific situation. Let $\mathbb{X}=L^1(\mathbb{R}_+,\mu)$ be the space of integrable functions with respect to a given measure $\mu$ and $F:L^1(\mathbb{R}_+,\mu)\to\mathbb{R}$ a linear functional defined as $$L(f):=\int_{\mathbb{R}_+}f(t)\,d\mu(t)$$ Let $A:=\{f\in L^1(\mathbb{R}_+,\mu), f\geqslant 0:||Tf||_{\infty}\leqslant 1\}$ where $T:L^1(\mathbb{R}_+,\mu)\to\text{ran}(T)$ is an integral operator defined as $$(Tf)(t):=\int^1_0f(tx)\,d\beta(x)$$ with $\beta(x)$ some given smooth function on $[0,1]$.

  • $\begingroup$ Without more context, we really don't know how to construct anything explicitly. We don/t even know any particular members of $A$. $\endgroup$ – Robert Israel Jul 27 '18 at 5:35
  • $\begingroup$ Perhaps you might start with precisely how $A$ is "given". Is it the set of all $x \in X$ satisfying a certain set of conditions? Assuming it is separable, is there a way to generate a sequence that is dense in $A$? $\endgroup$ – Robert Israel Jul 27 '18 at 13:11
  • $\begingroup$ @Arian: Why is the set $A$ in your $L^1$-example bounded? $\endgroup$ – Jochen Glueck Jul 27 '18 at 23:24
  • $\begingroup$ @JochenGlueck It is bounded for a given pair of $\beta$ and $\mu$ which I have not stated here (they both satisfy $\beta'(x)\geqslant 0,\mu'(x)\geqslant 0$). One then uses the asymptotic behavior of $\mu'(t)$ and the properties of functions $f\in A$ to get that the linear functional $L$ is uniformly bounded on $A$. Then use $L(f)=||f||_{L^1(\mathbb{R}_+,\mu)}$ for $f\geqslant 0$. $\endgroup$ – Arian Jul 28 '18 at 10:12

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