Delta function is not in $ L^1(\mathbb{R})$

Question

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?

Answer 1

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 $f.$

Answer 2

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*}

Answer 3

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.

Answer 4

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 $k > n$ $$\|h_k-h_n\|_{L^1} = \frac{1}{k} (k-n)+(\frac{1}{n}-\frac{1}{k}) n= 2-2\frac{n}{k}$$

Answer 5

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.