extension of linear functional from $C^0(\mathbb R)$ on $L^2(\mathbb R)$ I need to find out whether the linear functional on $C^0(\mathbb R)$ (set of continuous functions with compact support) can be extended to continuous linear functional on $L^2(\mathbb R)$.
1) for $\phi(f)=f(0)$
2) for $\phi(f)=\int_\mathbb Rf(x)dx$
3) for $\phi(f)=\int_\mathbb R f(x)g(x)dx$, where $g\in C^0(\mathbb R)$.
I think I'm done with first two. I would appreciate if someone checks me. Here is idea for 1): answer is no. Let's take 
$$f_n=\begin{cases}0, x\in(-\infty;\frac{-1}{n}]\\1+nx, x\in[\frac{-1}{n};0]\\1-nx, x\in[0;\frac{1}{n}]\\0, x\in[\frac{1}{n};\infty)
\end{cases}$$
Their $L^2$ norm equals to $\frac{2}{3n}$, so tends to $0$. Therefore, $f_n$ converges to $0$ in $L^2$, but $\phi(f_n)=1$ for all $n$.
For the 2) answer is yes. I only have to check that $\phi$ is continuous, because on the lecture we have proved theorem stating that every bounded functional $\phi: C^0\rightarrow \mathbb R$ can be extended to continuous $\bar{\phi}:L^2(\mathbb R)\rightarrow \mathbb R$, but on normed space continuity and boundedness are equivalent. So if $f_n\rightarrow f$, then for any $\epsilon>0$ there exists $N$ such that for any $x$ and for any $n>N$ we have $||f_n-f||_{L^2}<\epsilon$, which implies $\int_\mathbb R (f_n-f)^2dx<\epsilon$. Then |$\phi(f_n)-\phi(f)|=|\int_\mathbb R(f_n-f) dx|\le\int_\mathbb R|f_n-f|dx$. Now, using Cauchy-Schwarz inequality, we get $\int_\mathbb R|f_n-f|dx\le\sqrt{\int_\mathbb R (f_n-f)^2dx}<\sqrt{\epsilon}$. 
Am I right?
Can anyone help me with the third part, please?
Thanks.
 A: Your reasoning for 2) is flawed because what Cauchy-Schwarz implies is
$$
\int_{\Bbb R}|f_n-f|\le \sqrt{\int_{\Bbb R}1}\sqrt{\int_{\Bbb R}|f_n-f|^2}=\infty,
$$ but not $\int_{\Bbb R}|f_n-f|\le \sqrt{\int_{\Bbb R}|f_n-f|^2}$. This is because the Lebesgue measure on $\Bbb R$ is infinite. Now, to get an actual counterexample, let us consider any $g\in C^0(\Bbb R)$ with $g\ge 0$, $\ \int_{\Bbb R} g=1$. Let $g_n(x) =\frac1{\sqrt{n}}g(\frac{x}n)$. We can see
$$
\phi(g_n)=\frac1{\sqrt n}\int_{\Bbb R}g(\frac{x}n)\mathrm dx=\sqrt{n}\int_{\Bbb R} g=\sqrt n
$$ while $\|g_n\|^2_{L^2}=\frac1 n\int_{\Bbb R}|g(\frac x n)|^2\mathrm dx=\int_{\Bbb R}|g|^2$. If $\phi$ can be continuously extended to $L^2$, then there must be some $C>0$ such that for all $f\in C^0(\Bbb R)$, $|\phi(f)|\le C\|f\|_{L^2}$ holds. But then, $(g_n)_{n\ge 1}$ should satisfy that
$$
|\phi(g_n)|=\sqrt{n}\le C\|g_n\|_{L^2}=C\sqrt{\int_{\Bbb R}|g|^2}
$$ for all $n\ge 1$, which leads to a contradiction. So $\phi$ cannot be continuously extended to $L^2(\Bbb R)$.
A: For part 2, consider the function $f$ given by
$$f(x) = \frac{x}{1+x^2}.$$
Notice that $f$ decays like $\frac{1}{x}$ at $\pm\infty$.  This means two things:


*

*$f \in L^2(\mathbb{R})$, and

*$\int_\mathbb{R} f(x) \;dx = \infty$.


On the other hand, let $\chi_N$ be a cut-off function defined by
$$\chi_N(x) = \begin{cases}
  1 & |x| < \frac{N}{2}\\
  1 - \frac{2}{N}\left(|x| - N\right) & \frac{N}{2} \leq |x| \leq N\\
  0 & |x| > N
\end{cases}$$
Then, the sequence $f_n(x) = f(x) \chi_n(x)$ converges to $f$ in $L^2$ (by monotone convergence) with each $f_n \in C^0(\mathbb{R})$.  Thus, by monotone convergence, we would have
$$\phi(f) = \lim_{n\to\infty} \phi(f_n) = \int_{\mathbb{R}} f(x)\;dx = \infty$$
so $\phi$ cannot be extended to a continuous linear functional on $L^2$.

For part 3, try using Cauchy-Schwartz.  Can you show that $\phi$ is bounded?
