The Dirac delta is not in L^p

 
3 Theorem. Let $p ∈[1, ∞]$. Then there exists no $h ∈ L^p((0,2))$ such that \[ ∫_{(0,2)} h ⋅ φ ⋅ d λ^1=φ(1)   ∀ φ ∈ C_c^{∞}((0,2)) \] Proof. Assume there is an $h ∈ L^p((0,2))$ with the property above. Let $q$ be the conjugate of $p$. Further, let $φ_ϵ$ as in the lemma. Then we have \[ \begin{aligned} 1 & =\left|φ_ϵ(1)\right| \\ & =\left|∫ h φ_ϵ d λ^1\right| ⩽ ∫|h|\left|φ_ϵ\right| d λ^1=∫\left|h 1_{B_ϵ(1)}\right|\left|φ_ϵ\right| d λ^1 \\ & ⩽\| h 1_{B_ϵ(1)}\|_p ⋅\| φ_ϵ\|_q ⩽\| h 1_{B_ϵ(1)} \|_p \\ & =\left(∫\left|h 1_{B_ϵ(1)}\right|^p d λ^1\right)^{\frac{1}{p}} \xrightarrow{ϵ → 0+} 0 \end{aligned} \] (note the use of Hölder's inequality; the limit follows from the dominated convergence theorem which is applicable as \[ 0 \stackrel{\text { a.e. }}{⟵}\left|h 1_{B_ϵ(1)}\right|^p ⩽|h|^p \] and $|h|^p$ is integrable since $h ∈ L^p((0,2))$.) Contradiction!