Application of uniform boundedness principle

Question

$\DeclareMathOperator\Lip{Lip}$Let $\Lip_0(\mathbb R^d)$ be the space of Lipschitz functions $f:\mathbb R^d\to\mathbb R$ vanishing at zero, i.e., $f(0)=0$, and equipped with the norm $\|f\|:=\|\nabla f\|_{\infty}$. Following from Some natural subspaces and quotient spaces of $L^1$, $\big(\Lip_0(\mathbb R^d), \|\cdot\|\big)$ is a Banach space. Now we endow $\Lip_0(\mathbb R^d)$ with an alternative topology, denoted by $w$ and generated by the open sets $\mathcal O_{u}(f;\epsilon)$ as below:

$$\mathcal O_{u}(f;\epsilon) \quad:=\quad \left\{g\in \Lip_0(\mathbb R^d):~ \left|\int_{\mathbb R^d} \big[\nabla(f-g)(x)\cdot u(x)\big]\right| dx <\epsilon \right\},$$

where $f\in \Lip_0(\mathbb R^d)$, $u\in L^1(\mathbb R^d;\mathbb R^d)$ and $\epsilon>0$. My question is as follows: Let $(f_{\lambda})_{\lambda\in\Lambda}\subset \Lip_0(\mathbb R^d)$ be a net $w$-converging to $f\in \Lip_0(\mathbb R^d)$. Could we select a subnet $(f_{\lambda_{\alpha}})_{\alpha}$ s.t. $\sup_{\alpha}\|f_{\lambda_{\alpha}}\|<\infty$?

Answer 1

The answer is no, in general.

Before we discuss a counterexample, let us note that whenever a set $\mathcal{O}_u(f,\epsilon)$ contains $0$, then there is a another number $\tilde \epsilon > 0$ such that $\mathcal{O}_u(0,\tilde \epsilon) = \mathcal{O}_u(f,\epsilon)$. Indeed, $0 \in \mathcal{O}_u(f,\epsilon)$ implies that $$ \int_{\mathbb{R}^d} \nabla f \cdot u \; dx < \epsilon, $$ so $$ \tilde \epsilon := \epsilon - \int_{\mathbb{R}^d} \nabla f \cdot u \; dx $$ is a strictly positive number. Clearly, $\mathcal{O}_u(0,\tilde \epsilon) = \mathcal{O}_u(f,\epsilon)$.

The above argument shows that, in order to test whether a net $(f_\lambda)$ $\omega$-converges to $0$, it suffices the show that, for each $\epsilon > 0$ und each $u \in L^1(\mathbb{R}^d;\mathbb{R}^d)$, the net is eventually contained in $\mathcal{O}_u(0,\epsilon)$.

Now we can construct our

Counterexample. Let $d = 1$ and let $\mathcal{F}$ denote the set of all finite subsets of $L^1(\mathbb{R}; \mathbb{R})$; this set is directed with respect to set inclusion. For each $F \in \mathcal{F}$ we can find a function $h_F \in L^\infty(\mathbb{R}; \mathbb{R})$ such that

• $\|h_F\|_\infty \ge |F|$ and
• $\int_{\mathbb{R}} -h_F \cdot u \; dx < \frac{1}{|F|}$ for all $u \in F$.

Now define $g_F \in \operatorname{Lip}_0(\mathbb{R})$ by $$ g_F(x) = \int_0^x h_F(y) \; dy \qquad \text{for } x \in \mathbb{R}. $$ Then the net $(g_F)_{F \in \mathcal{F}}$ converges to $0$ with respect to the topology $\omega$ (by what we observed at the beginning of the post), but no subnet of $(g_F)_{F \in \mathcal{F}}$ is norm bounded.

Answer 2

I claim this is not an answer, but my understanding of the answer provided by Jochen Glueck. Indeed, for each $f\in {\rm Lip}_0(\mathbb R^d)$, we can define the map $T_f: L^1(\mathbb R^d;\mathbb R^d)\to\mathbb R$ by

$$T_f(u):=\int_{\mathbb R^d}\nabla f(x)\cdot u(x)dx.$$

As $\|T_{f}\|_{\infty}=\|\nabla f\|_{\infty} = \|f\|$, it suffices to show, in view of the uniform boundedness principle, that $\{T_{f_{\lambda}}(u)\}_{\lambda\in\Lambda}$ is bounded for every $u\in L^1(\mathbb R^d;\mathbb R^d)$.

Now we go back to the above example. With the choice $g_F$, one finds that

$$\left|\int_{\mathbb R^d} \nabla g_F(x)u(x)dx\right|\le \frac{1}{|F|},$$

which yields a convergent, thus bounded, subsequence $\{T_{g_{F_n}}(u)\}_{n\ge 1}$. Hence, $\{T_{g_{F_n}}\}_{n\ge 1}$ is the required subnet.