I have attempted the following part from exercise 5.49 in https://www.mat.uniroma2.it/~cannarsa/cam_0607.pdf :
- For every $n \in \mathbb{N}$, let $f_n \colon \mathbb{R} \to \mathbb{R}$ be defined by: $$ f_n(x) = \begin{cases} \frac{1}{2^n} &\text{if $x \in [2^n, 2^{n+1}]$} \\ 0 &\text{otherwise} \end{cases} $$
Show that:
- […]
- $\{f_n\}$ does not converge weakly in $L^1(\mathbb{R})$
Is the following proof correct? Can it be made more concise?
Aiming for a contradiction, suppose that $(f_n)$ does converge weakly in $L^1(\mathbb{R})$, i.e. there is $f \in L^1(\mathbb{R})$ such that $f_n \rightharpoonup f$.
By definition, this means that for all $\phi \in (L^1(\mathbb{R}))^*$, we have $\phi(f_n) \to \phi(f)$.
The dual of $L^1(\mathbb{R})$ contains $L^\infty(\mathbb{R})$ through the following embedding:
$$ \begin{aligned} L^\infty(\mathbb{R}) &\to (L^1(\mathbb{R}))^* \\ g &\mapsto \phi_{g} \end{aligned} $$
where $\phi_g$ is defined by:
$$ \phi_g(h) = \int_\mathbb{R} gh\,\mathrm{d}\mu $$
First we prove that $f$ is non-negative almost everywhere.
Indeed, aiming for a contradiction, suppose it is not.
This means there exists a measurable $A \subseteq \mathbb{R}$ with $f(x) < 0 $ for all $x \in A$.
Let $\chi_A$ denote the characteristic function of $A$, which is in $L^\infty(\mathbb{R})$, and consider $\phi_{\chi_A} \in (L^1(\mathbb{R}))^*$.
On the one hand we have:
$$\phi_{\chi_A}(f) = \int_\mathbb{R} \chi_A f\,\mathrm{d}\mu = \int_A f\,\mathrm{d}\mu < 0$$
On the other hand, for all $n$, since $f_n \ge 0$, we have:
$$\phi_{\chi_A}(f_n) = \int_\mathbb{R} {\chi_A}f_n\,\mathrm{d}\mu = \int_A f_n\,\mathrm{d}\mu \ge 0$$
So $\phi_{\chi_A}(f_n)$ cannot converge to $\phi_{\chi_A}(f)$ as $n \to \infty$, contradicting the assumption that $f_n$ converges weakly to $f$.
Now, we prove that $f$ must in fact be zero almost everywhere.
For any $m \in \mathbb{N}$, consider $g_m := {\chi_{[2^m,2^{m+1}]}} \in L^\infty(\mathbb{R})$ and the corresponding functional $\phi_{g_m} \in (L^1(\mathbb{R}))^*$.
Then:
$$\phi_{g_m}(f_n) = \int_{[2^m, 2^{m+1}]} f_n \,\mathrm{d}\mu = \begin{cases} \frac{1}{2^n} (2^{n+1}-2^n) = 1 &\text{if $n = m$} \\ 0 &\text{otherwise} \end{cases} $$
Hence $\phi_{g_m}(f_n) \to 0$ as $n \to \infty$.
But by hypothesis $\phi_{g_m}(f_n) \to \phi_{g_m}(f)$, so $\phi_{g_m}(f) = 0$.
That is:
$$\int_{[2^m,2^{m+1}]} f\,\mathrm{d}\mu = 0$$
Since we have shown that $f$ is non-negative almost everywhere, this implies it is in fact zero almost everywhere on $[2^m, 2^{m + 1}]$.
But this is true for all $m \in \mathbb{N}$, and since a countable union of null sets is still null, this means $f$ is zero almost everywhere on $\bigcup_{m \in \mathbb{N}} [2^m, 2^{m+1}] = [1, \infty)$.
Apply the same kind of argument with $\phi_{\chi_{(-\infty, 1]}}$ to show that $f$ is zero almost everywhere on $(-\infty, 1]$, concluding that $f$ is zero almost everywhere on all of $\mathbb{R}$.
So far we have proven that, assuming $f_n \rightharpoonup f$, we have that $f$ is zero almost everywhere.
But this leads to a contradiction.
Indeed, consider the constantly-1 function ${\chi_{\mathbb{R}}}$ and the corresponding functional $\phi_{\chi_\mathbb{R}} \in (L^1(\mathbb{R}))^*$, which is just integration (on all of $\mathbb{R}$).
Then:
$$\phi_{\chi_\mathbb{R}}(f_n) = \int_\mathbb{R} f_n\,\mathrm{d}\mu = \frac{1}{2^n}(2^{n+1}-2^n) = 1 \to 1 \text{ as $n \to \infty$}$$
However, since $f$ is zero almost everywhere:
$$\phi_{\chi_\mathbb{R}}(f) = \int_\mathbb{R} f\,\mathrm{d}\mu = 0$$
Thus $\phi_{\chi_\mathbb{R}}(f_n) \nrightarrow \phi_{\chi_\mathbb{R}}(f)$, contradicting our initial assumption of the weak convergence of $(f_n)$ to $f$.
Best Answer
More concise. We still need the part about identifying $L^\infty$ with functionals on $L^1$.
Let $g \in L^\infty$ be defined by: $$ g(x) = (-1)^n\quad\text{for } x \in (2^n,2^{n+1}] $$ where $n=1,2,3,\dots$ and $g(x) = 0$ elsewhere. Then $$ \int_\mathbb R f_n g \;d\mu= (-1)^n,\quad n=1,2,3,\dots $$ This is not a Cauchy sequence of real numbers, so $\int_\mathbb R f_n g \;d\mu$ does not converge to anything. In particular, it does not converge to $\int_\mathbb R f g \;d\mu$ for any function $f$.