Skip to main content
edited body
Source Link
Falrach
  • 4.3k
  • 17
  • 40

Let $g \in L^{2}$ and $\epsilon >0$. There exists $\phi \in C_c^{\infty} (\mathbb R)$ such that $\|g-\phi\|_2 <\epsilon$. Hence $|\int f_n g|\leq |\int f_n \phi|+\|f_n\|\epsilon$ (by Holder'sHölder's/ C-S inequality). It is clear now that $\int f_n g \to 0$.

Let $g \in L^{2}$ and $\epsilon >0$. There exists $\phi \in C_c^{\infty} (\mathbb R)$ such that $\|g-\phi\|_2 <\epsilon$. Hence $|\int f_n g|\leq |\int f_n \phi|+\|f_n\|\epsilon$ (by Holder's/ C-S inequality). It is clear now that $\int f_n g \to 0$.

Let $g \in L^{2}$ and $\epsilon >0$. There exists $\phi \in C_c^{\infty} (\mathbb R)$ such that $\|g-\phi\|_2 <\epsilon$. Hence $|\int f_n g|\leq |\int f_n \phi|+\|f_n\|\epsilon$ (by Hölder's/ C-S inequality). It is clear now that $\int f_n g \to 0$.

Source Link
Kavi Rama Murthy
  • 361.2k
  • 20
  • 102
  • 195

Let $g \in L^{2}$ and $\epsilon >0$. There exists $\phi \in C_c^{\infty} (\mathbb R)$ such that $\|g-\phi\|_2 <\epsilon$. Hence $|\int f_n g|\leq |\int f_n \phi|+\|f_n\|\epsilon$ (by Holder's/ C-S inequality). It is clear now that $\int f_n g \to 0$.