# Application of uniform boundedness principle

$$\DeclareMathOperator\Lip{Lip}$$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. $$\sup_{\alpha}\|f_{\lambda_{\alpha}}\|<\infty$$?

• @NateEldredge Thanks for the remark. Of course a convergent net is not bounded, e.g. math.stackexchange.com/questions/3115924/… I've edited my question – Neymar Dec 27 '19 at 23:37
• May I ask why you care about this topology? The most natural topology on ${\rm Lip}_0(\mathbb{R}^d)$ is the weak* topology, which agrees with the one you describe on bounded sets. Maybe it would work better? – Nik Weaver Dec 28 '19 at 4:25
• @NikWeaver This is related to my previous posts mathoverflow.net/questions/346702/… and mathoverflow.net/questions/346680/… Yes. I agree this topology, restricted on bounded sets is metrizable, while in general is not. That's why I post it here. For a general converging net, could we always select a bounded subnet? – Neymar Dec 28 '19 at 4:49
• @NikWeaver Could you please specify the weak* topology that you mentioned? – Neymar Dec 28 '19 at 4:53
• See my book Lipschitz Algebras, second edition. – Nik Weaver Dec 28 '19 at 14:09

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.

• I don't get the the convergence of $(g_F)_{F\in\mathcal F}$ to $0$. Could you please explain a bit more? See e.g. my question below for details. – Neymar Dec 28 '19 at 14:29
• Could you point out if my arguments below are wrong? – Neymar Dec 28 '19 at 14:43

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.