Continuity of Function Related to $F$-norms

Question

Let $X$ be a locally bounded $F$-space and $\left\|\cdot\right\|$ be an $F$-norm on $X$. Suppose that $\left\|\cdot\right\|$ is concave: for all $x\in X$ fixed, the function $t\mapsto\left\|tx\right\|$ is concave on $(0,\infty)$.

For $r>0$, define $$ n(r):=\sup_{\left\|2x\right\|\leq r}\left\|x\right\|$$

It is not hard to show that for $x$ fixed, the function $t\mapsto\left\|tx\right\|$ is non-decreasing so that $n(r)\leq r$. In fact the inequality is strict, as a consequence of local boundedness.

I am reading the paper "A Generalization of Mazur-Ulam" by S. Rolewicz and in the proof of Lemma 2, author claims $n(r)$ is "trivially continuous", but I don't see that it is continuous. I cannot come up with a counterexample and the proof of Lemma 2 relies on the continuity, so I am pretty sure that it is true and I am just embarrassingly missing something simple.

I tried playing around with the quantity $\sup_{\left\|2x\right\|\leq r}[\left\|(1+t)x\right\|-\left\|x\right\|]$, for $t>0$, and the concavity of the norm, but I have not gotten far.

Would someone please direct me as to what I am missing?

Answer 1

Due to a misunderstanding on my behalf my original answer was flawed. My new argument does not fully work (I have indicated the gap), but as I think this is better than nothing, I'll leave it.

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Let $r_0$ be as in the cited paper. Then every nonzero vector in $X$ has a multiple with norm at least $2r_0$. It suffices to show (for the purposes of the paper) that $n|_{(0,r_0)}$ is continuous. We will from now on restrict $n$ to this interval.

It suffices to show that $n$ is convex, since convex functions are always continuous. Now let $v\mathbb R\subset X$ denote the one dimensional subspace spanned by $v\in X\setminus\{0\}$ and define (for $r<r_0$) $$ n_v(r)=\sup_{x\in v\mathbb R\atop \|2x\|\leq r}\|x\|. $$ Note that $n(r)=\sup_{v\in X\setminus\{0\}}n_v(r)$. Since the supremum of convex functions is convex, it suffices to show that each $n_v$ is convex.

We only have to solve our problem in one dimension! Let $\phi:[0,\infty)\to[0,\infty)$ be an increasing concave function with $\phi(0)=0$. The concave norm on our one dimensional space is of the form $x\mapsto\phi(|x|)$ for some such $\phi$. We also know that $\lim_{t\to\infty}\phi(t)\geq2r_0$.

So let us define $$ m(r)=\sup_{t\geq0\atop\phi(2t)=r}\phi(r) $$ for $r<r_0$. Our goal is to prove that $m$ is convex.

Let $T=\sup\{t>0;\phi(t)<2r_0\}$. We eventually only need to consider $\phi$ on $[0,T)$ due to our restriction of $n$. Suppose $\phi$ were not strictly increasing on this interval. Then there would be some $\tau<T$ so that $\phi$ is constant on $[\tau,T]$. But $\phi$ is concave on all of $[0,\infty)$, so this would mean that $\phi(t)=\lim_{t\to\infty}\phi(t)\geq2r_0$ for all $t\geq\tau$, contradicting the definition of $T$. Thus $\phi|_{[0,T)}$ is strictly increasing.

Now $\phi$ has an inverse function $\alpha:[0,2r_0)\to[0,T)$ and $m(r)=\phi(\frac12\alpha(r))$. Note that now $\alpha$ is strictly increasing and convex with $\alpha(0)=0$. It suffices to show that $m$ is convex, given this special structure. I just learned that this is not generally true, so this argument does not finish the proof.

(I'm not sure if the author of the paper had an easier argument in mind. It could also be that the claim seems too plausible to be false.)

Answer 2

Following an argument of Wobst, we can prove Lemma 2 without proving that $r\mapsto n(r)$ is continuous on $(0,r_{0})$. Instead, we prove the inequality

$$s_{r;r_{0}}:=\inf_{r\leq\left\|x\right|\leq r_{0}}\dfrac{\left\|2x\right\|}{\left\|x\right\|}>1,$$ where of course $0<r<r_{0}$, by using the concavity of $t\mapsto\left\|tx\right\|$ and the boundedness of the set $K_{2r_{0}}=\left\{x\in X:\left\|x\right\|\leq 2r_{0}\right\}$. It may be the case that this inequality implies the continuity of $n(r)$, but this implication is not yet evident to me.

Fix $x\in X$. By the concavity of the norm, we have $$\left\|2x\right\|=\left\|\frac{n+1}{n}x+\frac{n-1}{n}x\right\|\geq\frac{1}{n}\left\|(n+1)x\right\|+\frac{n-1}{n}\left\|x\right\|$$ Writing $\left\|2x\right\|=(1+\varepsilon_{x})\left\|x\right\|$, where $\varepsilon_{x}\geq 0$ ($t\mapsto\left\|tx\right\|$ is nonincreasing), we obtain the inequality $$\left\|(n+1)x\right\|\leq (1+n\varepsilon_{x})\left\|x\right\|$$

Suppose $s_{r;r_{0}}=1$. Then there exists a sequence of points $(x_{k})$ with $r\leq\left\|x_{k}\right\|\leq r_{0}$ and $\left\|2x_{k}\right\|/\left\|x_{k}\right\|\rightarrow 1$. Define $\varepsilon_{k}\geq0$ by $\left\|2x_{k}\right\|=(1+\varepsilon_{k})\left\|x_{k}\right\|$. Observe that $\varepsilon_{k}>0$ for all $k$, otherwise the boundedness of $K_{2r_{0}}$ implies the sequence $n^{-1}\cdot(nx_{k})$ tends to $0$. We will arrive at a contradiction by showing that $\varepsilon_{k}$ cannot tend to $0$.

For each $k$, let $n_{k}$ be the maximal integer such that $\varepsilon_{k}\leq n_{k}^{-1}$. Note that $n_{k}\rightarrow\infty$ as $k\rightarrow\infty$. I claim that the sequence $((n_{k}+1)x_{k})$ is contained in $K_{2r_{0}}$. Indeed,

$$\left\|(n_{k}+1)x_{k}\right\|\leq(1+n_{k}\varepsilon_{k})\left\|x_{k}\right\|\leq 2\left\|x_{k}\right\|\leq 2r_{0}$$

Then the sequence $(n_{k}+1)^{-1}\cdot(n_{k}+1)x_{k}$ tends to $0$, which is a contradiction.

Let $(r_{n})_{n=0}^{\infty}$ be the decreasing sequence defined in Rolewicz's paper, and suppose that $\lim r_{n}=r'>0$. Then

$$r'=\lim_{n\rightarrow\infty}n(r_{n})\leq\dfrac{\lim_{n\rightarrow\infty} r_{n}}{s_{r';r_{0}}}=\dfrac{r'}{s_{r';r_{0}}}<r',$$ a contradiction.