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I am trying to prove the following:

If $X$ is a finite-dimensional space, then for sequences $\left\{x_n\right\}\subseteq X$ and $\left\{f_n^*\right\}\subseteq X^*$, if there exists an $x\in X^*$ such that $x_n \rightharpoonup x$ and $f_n\stackrel{*}{\rightharpoonup} f$, then we have $x_n\rightarrow x$ and $f_n \rightarrow f$.

I already figured out how to prove that weak convergence implies regular convergence:

Suppose $x_n \rightharpoonup x$ and that $X$ is of finite dimension $m$. Let $\left\{e_1, \ldots, e_m\right\}$ be the basis of $X$. Then we can write any $x_n=\displaystyle \sum_{i=1}^m c_{i,n}e_i$ and $x=\displaystyle \sum_{i=1}^m c_ie_i$, where $c_{i,n}, c_i \in \mathbb{R}$. Let $\left\{f(e_1), \ldots , f(e_m)\right\}$ be the basis of $X^*$. Since $f(x_n)\rightarrow f(x)$ in $X^*$, this implies that $$\left\|f(\sum_{i=1}^m c_{i,n}e_i)-f(\sum_{i=1}^m c_ie_i)\right\|=\left\|f(\sum_{i=1}^m (c_{i,n}-c_i)e_i) \right\| \rightarrow 0,$$

which is true if and only if $c_{i,n}\rightarrow c_i$ for all $i\in \left\{1, \ldots, m\right\}$. Thus, $\left\|x_n-x\right\|=\left\|\sum_{i=1}^m (c_{i,n}-c_i)e_i\right\|\leq \sum_{i=1}^m |c_{i,n}-c_i|\left\|e_i\right\| \rightarrow 0$, that is, $x_n \rightarrow x$. \

I don't know how to show that $f_n\stackrel{*}{\rightharpoonup} f$ implies $f_n \rightarrow f$ and I think it's mostly because I don't see the difference between the two. Can someone show me how to prove this? Thank you!

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The notations you use are a bit confusing. For example saying $\{f(e_1),...,f(e_n)\}$ is a basis of $X^*$ is false, since $f$ is a functional defined on $X$, and therefore $f(e_i)$ are real numbers, not functions.

$x_n$ converges weakly to $x$ if and only if $f(x_n) \to x$ for every $f \in X^*$.

$f_n$ converges weak * to $f$ if and only if $f_n(x) \to f(x)$ for every $x \in X$.

You want to prove that in finite dimensional spaces weak convergences are strong. If $x_n$ converges weakly to $x$, then pick the functional $f_i$ which associates to $x$ the $i$-th coordinate, i.e. $f_k(\sum c_{i,n}e_i)=c_{k,n}$. You have $f_i(x_n) \to f_i(x)$ for every $i$. Therefore the coordinates of $x_n$ converge to the coordinates of $x$. The last inequality you wrote proves that $\|x_n-x\| \to 0$.

If $f_n$ converges weakly * to $f$ then $f_n(x) \to f(x)$ for every $x$. You have $$ \|f_n-f\| = \sup_{\|x\|\leq 1} |f_n(x)-f(x)| \leq \sum_{i=1}^n |f_n(e_i)-f(e_i)| \to 0$$ because we have a sum of $n$ sequences which converge to zero.

I used the fact that $$|f_n(x)-f(x)| \leq \sum |c_i| |f_n(e_i)-f(e_i)|$$ and each $c_i \leq 1$ since $\|x\|\leq 1$.

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