This makes me puzzled.Here $H^{k}(\Omega)$ is the sobolev space $W_p^k(\Omega)$ with $p=2$.
We all know that $H^{k}(\Omega)$ is Hilbert space,This means that $(H^k(\Omega))^{*}=H^{-k}(\Omega)$ should be $H^{k}(\Omega)$ itself from Risez Representation theorem.
However,we know that Dirac $\delta$-function $\delta\in W^{k}_p(\Omega)$ if $k<-n+n/p \ $ from Sobolev's Inequality.Clearly $\delta\notin H^{k}(\Omega),\forall k>0$.This will be a contradiction,cause there are some $k$ such that $\delta\in H^{-k}(\Omega)$ while $\delta\notin H^{k}(\Omega)$.
What's wrong with my previous statement?
This is a good question. A while ago, I was puzzled by this myself.
The apparent contradiction is resolved by noting that this is a problem of choosing the right "dual pairing".
You have indeed $(H^s)^\ast = H^s$, if you identify a function $f \in H^s$ with the functional $g \mapsto \langle g, f\rangle_{H^s} = \sum_{|\alpha|\leq s} \int \partial^\alpha g \overline{\partial^\alpha f}$ (or whatever scalar product you use on $H^s$).
On the other hand, when you write $\delta \in H^{-s}$, then you mean the functional $f \mapsto f(0)$. Essentially, what one uses here is the dual pairing from the (tempered) distributions, which extends the pairing $\langle f,g\rangle = \int f \overline{g}$.
The (perhaps surprising) consequence of this is that (by the Riesz representation theorem) there is a function $g \in H^s$ with $\delta(f) = \langle f,g\rangle_{H^s}$ for all $f \in H^s$.