Prove that: $\delta_{1}-\delta_{-1} \notin L^{2}(\mathbb{R})$

Question

I'm going to prove that there is no $f\in L_2(\mathbb{R})$ such that

$$\langle \delta_{1}-\delta_{-1},\varphi\rangle =\int f\varphi $$

for all test functions $\varphi$ (i.e. $\varphi\in \mathcal{D}$), where $$\delta_a= \operatorname{Distribution Dirac}\quad\text{at}\quad a$$ that is $$\langle \delta_{a},\varphi\rangle = \varphi(a)$$ for any $a\in\mathbb{R}$.

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My attempt:

Suppose $\delta_{1}-\delta_{1} \in L^{2}(\mathbb{R})$

$$\langle \delta_{1}-\delta_{1} ,\varphi \rangle = \varphi(1)-\varphi(-1)$$ Mean! $$\exists c>0 , |\varphi(1)-\varphi(-1)|≤c \|\varphi\|_{L^{2}} $$ Here I need find a sequence of continuous functions to get a contradiction!

Can you assist? I'm grateful!

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Thanks!

Answer 1

What you really mean to ask is this. Show there doesn't exist a function $f \in L^2(\mathbb R)$ such that for any test function $h$ $$ (\delta_{1} - \delta_{-1})(h) := h(1) - h(-1) = \int_{\mathbb R} f(x) h(x) \, dx .$$ Suppose there does exist such an $f$. If $h$ is any test function, find a sequence of test functions $h_n$ for which $h_n(1) = h_n(-1) = 0$ such that ${\|h - h_n\|}_2 \to 0$. Thus $$ \int _{\mathbb R} f(x) h(x) \, dx = 0 ,$$ and we have a contradiction.