For Sobolev spaces $H^s(R^d)$, with $s> \frac{d}{2}$ every element of $H^s(R^d)$ is an equivalence class $[f]$ and in every such a class there exists a unique continuous function $f^{*}$. Can we define the trace operator $tr_{R^{d_0}}: H^s(R^d)\to H^{s-\frac{d-d_0}{2}}(R^{d_0})$, on some $R^{d_0}$ with $d_0<d$ as:

$\hspace{3cm}tr_{R^{d_0}}[f]=[f^{*}|_{R^{d_0}}]$ $\ \ \ $? $\ \ \ $ What is wrong with this definition ?

As far as I know the trace operator is defined as a continuous extension of the restriction operator, initially defined from $C^{\infty}_c(R^d)$ to $C^{\infty}_c(R^{d_0})$

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