Let $X$ be a Banach space and denote by $B(x,r)$ the closed ball centered at $x\in X$ and radius $r>0$. We recall that the Hausdorff measure of noncompactness (MNC) is defined as $$ \beta(B):=\inf\{r>0 : B\subset \cup_{i=1}^{n}B(x_{i},r),\textrm{ for some } x_{1},\ldots ,x_{n}\in X\}. $$ for each non-empty and bounded $B\subset X$. The Kuratowski MNC is defined as $$ \alpha(B):=\inf\{r>0 : B\subset \cup_{i=1}^{n}D_{i},\textrm{ for some } D_{1},\ldots ,D_{n}\subset X, diam(D_{i})\leq r\}. $$ where $diam(D_{i})$ is the diameter of $D_{i}$. For instance, if $X$ is the space of the continuos functions defined on $[0,1]$, for the set $$ C:=\big\{f\in X: 0\leq f(0)\leq \frac{1}{3},0\leq f(x)\leq 1, \frac{2}{3}\leq f(1)\leq 1 \big\}, $$ I have read, without proof that $\alpha(C)=1$ and $\beta(C)=1/2$.
Now, assume $X$ is a Hausdorff locally convex topological vector space (LCTVS), but not a Banach space, whose topology is generated a family of seminorms $P$. For each $p\in P$, we denote by $B_{p}(x,r):=\{y\in X:p(x-y)\leq r\}$ the closed "ball" defined by $p$. We can define in the class of bounded subsets of $X$ the function $$ \beta_{p}(B):=\inf\{r>0 : B\subset \{x_{1},\ldots,x_{n}\}+B_{p}(0,r),\textrm{ for some } x_{1},\ldots ,x_{n}\in X\}. $$ Replacing, in the above definition of $\alpha$, $diam(D)$ by $diam_{p}(D):=\sup\{p(x-y):x,y\in D\}$ we obtain $$ \alpha_{p}(B):=\inf\{r>0 : B\subset \cup_{i=1}^{n}D_{i},\textrm{ for some } D_{1},\ldots ,D_{n}\subset X, diam_{p}(D_{i})\leq r\}. $$
Thus, the families $(\beta_{p})_{p\in P}$ and $(\alpha_{p})_{p\in P}$ are called, respectively, the Hausdorff and the Kuratowski MNC.
I am looking for an example where $$ 0<\beta_{p}(C)\neq \beta(C) \textrm{ and/or } 0<\alpha_{p}(C)\neq \alpha(C) \quad (*) $$ for some closed and convex $C$ subset of $X$. My attempts have been the following.
Let $X$ be the Banach space of the continuos functions defined on $[0,1]$, and $C$ as above. Consider $X$ endowed the point-wise convergence topology. Such topology is generated by the seminorms $p_{x}(f):=|f(x)|$ for each $x\in [0,1]$. As $[0,1]$ is compact, given $\varepsilon>0$ there is a finite set, put $F=\{x_{1},\ldots,x_{n}\}\subset [0,1]$ such that for each $x\in[0,1]$ there exists $x_{i}\in F$ such that $|x-x_{i}|\leq \varepsilon$. Then, by putting $f_{i}(x):=x_{i}$ for each $x\in [0,1]$, for each $x\in [0,1]$ for a given $f\in C$ there is $f_{i}$ such that $$ p_{x}(f-f_{i})=|f(x)-f_{i}(x)|=|f(x)-x_{i}|\leq \varepsilon. $$ So, $\beta_{p_{x}}(C)=0$. Likewise, $\alpha_{p_{x}}(C)=0$ for each $x\in [0,1]$. I am not sure if the above proof is correct at all, but if it is correct, (*) is not satisfied.
If $X$ is the space of the null sequences $c_{0}$ endowed the topology generated by the seminorms $p_{n}(x):=|x_{n}|$ for each $n\geq 1$ and $x:=(x_{n})_{n\geq 1}\in c_{0}$, I think (reasoning as above) that $\beta_{p_{n}}(B)=\alpha_{p_{n}}(B)=0$ for each $n\geq 1$, where $B$ stands for the unit closed ball of $c_{0}$. So, (*) is not satisfied in this case. Note that $\beta(B)=1$ and $\alpha(B)=2$ if $c_{0}$ is endowed its usual supremum norm.
Somebody know an example where (*) be satisfied?
Many thanks for your comments and suggestions.
We can trivially provide the required examples as follows. Let $(X,\|\cdot\|)$ be any Banach space. For any $\lambda>0$ and $x\in X$ put $p_\lambda(x)=\lambda\|x\|$. Then $p_\lambda$ is a norm on $X$. We have $B_{p_\lambda}(x,r)=B(x,r/\lambda)$ for each $x\in X$ and $\operatorname{diam}_{p_\lambda}(D)=\lambda\operatorname{diam}(D)$ for any nonempty subset $D$ of $X$. It follows that $\beta_{p_\lambda}(B)=\lambda\beta(B)$ and $\alpha_{p_\lambda}(B)=\lambda\alpha(B)$ for any nonempty bounded subset $B$ of $X$. In particular, for the closed convex set $C$ from the question and any $\lambda\ne 1$ we have $0<\beta_{p_\lambda}(C)=\lambda\beta(C)\ne \beta(C)$ and $0<\alpha_{p_\lambda}(C)=\lambda\alpha(C)\ne \alpha(C)$.