Delta function is not in $ L^1(\mathbb{R})$ Let $f_n$ be a sequence in $L^1(\mathcal{R})$ such that
$$
\lim_{n\to \infty} \int_{\mathbb{R}}f_n(x)g(x)dx = g(0)
$$
for each $g\in C_0(\mathbb{R})$.  Show $f_n$ is not Cauchy.
My approach was to argue by contradiction and use that in $L^1$ Cauchy and Convergence are equivalent.  So I assume $f_n \to f$ in $L^1$ and then we get that $\int fg = g(0)$ for all $g$ vanishing at infinity.  Then then inequality
$$
g(0) = \int fg \leq \max |g|\int|f|
$$
which yields $1\leq \| f\|_1$.  I would like to find an upper bound for the 1-norm of $f$ which is smaller than 1 so that I get a contradiction.  But I do not know how to approach this.  How should I proceed?
 A: If $f$ were $L^1$ then consider a sequence $g_n$ with$g_n(0)=1$ and $\operatorname{supp}g = (-1/n,1/n)$ and $\|g_n\|_{C^0} =1$. By Dominated Convergence
, since $fg_n\to 0$,
$$1=g_n(0) = \int fg_n \to 0 $$
Which is absurd. 
A: If there were such an $f,$ we would have
$$\tag 1\int_{-\infty}^\infty f(x)\frac{\cos nx}{1+x^2}\, dx  =1,\,\, n=1,2,\dots$$
But $f(x)/(1+x^2) \in L^1,$ and thus $(1)\to 0$ as $n\to \infty$ by the Riemann-Lebesgue lemma. That's a contradiction, proving there is no such $f.$ 
A: We contend that $\displaystyle\int f(x)g(x)dx=g(0)$ for every $g\in C_{0}$ will lead to a contradiction, here $f$ is the $L^{1}$ limit of $(f_{n})$. Let $\epsilon>0$ so small such that $\displaystyle\int_{|x|\leq 2\epsilon}|f(x)|dx<\dfrac{1}{2}$ and let $g\in C_{0}$ be such that $0\leq g(x)\leq 1$, $g(x)=1$ for $|x|\leq\epsilon$, $g(x)=0$ for $|x|\geq 2\epsilon$. Then
\begin{align*}
1=|g(0)|=\left|\int f(x)g(x)dx\right|\leq\int_{|x|\leq 2\epsilon}|f(x)|dx<\dfrac{1}{2}.
\end{align*}
A: For any $g \in L^1$, let $$g_n(x) = g \ast n 1_{|x| < \frac{1}{2n}}(x) = \int_{\frac{-1}{2n}}^{\frac{1}{2n}} n g(x+y)dy$$ Then $g_n \to g$
in $L^1$.
If your sequence $f_m$ is Cauchy in $L^1$ then the limit is $\in L^1$ thus the sequence
$$h_n =\lim_{m \to \infty} f_m \ast n 1_{|x| < \frac{1}{2n}}=n 1_{|x| < \frac{1}{2n}}$$
is Cauchy in $L^1$.
But for $k > n$ $$\|h_k-h_n\|_{L^1} = \frac{1}{k} (k-n)+(\frac{1}{n}-\frac{1}{k}) n= 2-2\frac{n}{k}$$
A: Suppose ${f}_{n}$ is a Cauchy sequence. For $n$ and $m$
large enough, we have ${\left\|{f}_{n}-{f}_{m}\right\|}_{{L}^{1} \left(\mathbb{R}\right)}  \leqslant  \frac{1}{3}$
Let ${g}_{a} \left(x\right) = {e}^{{-a} {x}^{2}}\in {\cal C}_0({\mathbb R})$. For a given $n$, choose $a$
large enough so that
$$\left|\int_{}^{}{f}_{n} {g}_{a} d x\right|  \leqslant  \frac{1}{3}$$
We have, for all $m$ large enough
$$\left|\int_{}^{}{f}_{m} {g}_{a} d x\right|  \leqslant  \left|\int_{}^{}{f}_{n} {g}_{a} d x\right|+{\left\|{f}_{n}-{f}_{m}\right\|}_{{L}^{1} \left(\mathbb{R}\right)}  \leqslant  \frac{2}{3}$$
Letting $m \rightarrow  \infty $ we get $1  \leqslant  \frac{2}{3}$, a contradiction.
