For each $ n \in \mathbb{N} $ let $ T_n : L^{\infty}( (0, + \infty)) \to \mathbb{R} $ be defined as $$ T_n(f) = n \Biggl ( \int_0^1 x^n f(x) dx + \int_1^{+ \infty}e^{-nx}f(x)dx \Biggr )$$

Does there exist a functional $T : L^{\infty}( (0, + \infty)) \to \mathbb{R} $ s.t. $T_n \stackrel{*}{\rightharpoonup} T$?

I was able to prove that $T_n \to \delta_1$ in the sense of distributions and also as a functional on the bounded and continuous functions defined on $(0, + \infty)$.

What can I do for the whole $L^{\infty}( (0, + \infty))$?


The sequence $T_n$ does not have a weak* limit in $(L^\infty)^*$. (Of course since the norms are bounded it has weak* limit points.)

Writing down all the details would take some time and space, and result in something that would be tedious to read. Here's an indication of how the construction goes; you can fill in the details:

First, for convenience define $$\phi_n(t)=(n+1)t^n\chi_{(0,1)}(t),$$so that $\int \phi_n$ is exactly $1$. We will show that there exists $f\in L^\infty$ and a sequence $n_j\to\infty$ so that $$\int \phi_{n_j}(t)f(t)\sim(-1)^j.$$We can take $f$ to vanish off $(0,1)$, and then it follows that $$T_{n_j}f\sim(-1)^{j}.$$

Suppose that $a\in(0,1)$ and $\epsilon>0$ are given. There exists $n$ so that $$\int_a^1\phi_n(t)>1-\epsilon,$$and now there exists $b\in(a,1)$ so that $$\int_a^b\phi_n(t)>1-\epsilon.$$

It follows that we can recursively define a sequence $a_j$ increasing to $1$ and a sequence $n_j\to\infty$ so that $$\int_{a_j}^{a_{j+1}}\phi_{n_j}(t)\to1.$$Which of course is the same as$$\int_{(0,1)\setminus(a_j,a_{j+1})}\phi_{n_j}(t)\to0.$$

Now say $I_j=(a_j,a_{j+1})$ and set $$f=\sum f_j=\sum(-1)^j\chi_{I_j}.$$ Then $$\int \phi_{n_j}(t)f(t)\sim(-1)^j.$$Proof: It's clear that $$\int \phi_{n_j}(t)f_j(t)\sim(-1)^j$$and $$\left|\int \phi_{n_j}(t)(f(t)-f_j(t))\right|\le\int_{(0,1)\setminus I_j}\phi_{n_j}\to0.$$

Note of course the sequence $T_n$ has a weak* convergent subnet. But it does not have a weak* convergent subsequence: Given any sequence $n_j$ increasing to $\infty$ the constructon above gives a subsequence $n_{j_k}$ and $f\in L^\infty$ such that $T_{n_{j_k}}f\sim(-1)^k$.


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