# $Weak^*$ convergence in $L^1$

Suppose we have a sequence $$\{f_n\}$$ of $$L^1$$ functions such that $$||f_n||_1 \leq K_1$$, then viewing $$L^1(\mathbb{R}) \subset \mathcal{M}(\mathbb{R})$$ where $$\mathcal{M}(\mathbb{R})$$ is the space of Radon measure which is isomorphic to the dual space of $$C_C(\mathbb{R})$$, we can extract a subsequence $$\{{f_n}_k\}$$which converges to a Radon measure in the $$weak^*$$ topology on $$\mathcal{M}(\mathbb{R})$$. Suppose in addition we have

$$\int_{\mathbb{R}}f_n=K_2, \forall n \in \mathbb{N}$$ and $$f_n \rightarrow f$$ pointwise almost everywhere in $$\mathbb{R}/\{0\}$$

then can we say that $$\exists C \in \mathbb{R}$$ such that $$\int_{\mathbb{R}} {f_n}_k\Phi(x)dx=\int_{\mathbb{R}} f(x)\Phi(x)dx+ C\Phi(0)$$

i.e the $$weak^*$$ limit is of the subsequence is of the form $$f+C\delta_{0}$$

Note that convergence almost everywhere in $$\mathbb R\setminus\{0\}$$ is equivalent to convergence almost everywhere in $$\mathbb R$$ since $$\{0\}$$ has measure zero.
Now let $$f_n = n \chi_{(1,1+1/n)}$$. It converges pointwise a.e. to zero, weak star to $$\delta_1$$.
• Thanks.. So the limit is not always of the form $f+C \delta_0$, but can be of the form $f+C \delta_{x_{0}}$ where $x_0 \in \mathbb{R}$ as well. Is this always true? I mean whenever the sequence satisfies the conditions mentioned in my question can we say that the $weak^*$ limit is always of the form $f+C \delta_{x_{0}}$ – Rosy Feb 13 at 15:58
• Suppose in addition we have $f_n \rightarrow f$ in $L^1(\mathbb{R}/B(0,\epsilon))$ for all $\epsilon>0$ then can we say that $\exists C \in \mathbb{R}$ such that $\int_{\mathbb{R}} {f_n}_k\Phi(x)dx \rightarrow \int_{\mathbb{R}} f(x)\Phi(x)dx+ C\Phi(0)$ i.e the $weak^*$ limit in $\mathbb{R}$ of the subsequence is of the form $f+C\delta_{0}$ – Rosy Feb 13 at 16:21