Consider two different norms on the Sobolev space $W^{1,2}([0,1])$: the $L_2$ norm $$ \|f\|_{L_2}^2:=\int_{[0,1]} f^2(x)\, dx $$ and the usual standard norm:

$$ \|f\|_{H^1}^2:=\|f\|_{L_2}^2+\|f'\|_{L_2}^2. $$

It is obvious that this two norms are not equivalent and generate different topologies. Is it true that these two norms, nevertheless, generate the same Borel $\sigma$--algebra?

  • 1
    $\begingroup$ How do norms generate a sigma algebra? $\endgroup$
    – daw
    May 23, 2018 at 20:32
  • 1
    $\begingroup$ @daw thanks for your comment. I missed the word Borel $\sigma$--algebra. $\endgroup$
    – Oleg
    May 23, 2018 at 20:56

1 Answer 1


This is indeed true.

Since the $\sigma$-algebra generated by $\|\bullet\|_{H^1}$ is clearly finer than the one generated by the $L^2$-norm, it suffices that $\|\bullet\|_{H^1}$ is measurable with respect to the $L^2$-Borel-$\sigma$-algebra. For this, it suffices to show that $f \mapsto \|f'\|_{L^2}$ is measurable.

To show this, let $\Gamma \subset C_c^\infty ((0,1)) \setminus \{0\}$ be countable and such that $\Gamma \subset L^2$ is dense. Since $L^2$ is separable, and since $C_c^\infty$ is dense in $L^2$, such a set exists. It is not hard to see for any $h \in L^2$ that $$ \|h\|_{L^2} = \sup_{g \in \Gamma} \frac{|\langle h, g \rangle|}{\|g\|_{L^2}} \, . $$

If we apply this to $f' \in L^2$ for $f \in H^1$, we see $$ \|f'\|_{L^2} = \sup_{g \in \Gamma} \theta_g (f) \, , $$ with $$ \theta_g (f) := \frac{|\langle f', g \rangle|}{\|g\|_{L^2}} = \frac{|\langle f, g' \rangle|}{\|g\|_{L^2}} \, . $$ This identity show that $H^1 \to \Bbb{R}, f \mapsto \theta_g (f)$ is continuous (and hence measurable) with respect to $\|\bullet\|_{L^2}$.

As a countable supremum of measurable functions, the function $H^1 \to \Bbb{R}, f \mapsto \|f'\|_{L^2}$ is thus measurable.

  • $\begingroup$ Wow! that's a very nice argument, thanks a lot! $\endgroup$
    – Oleg
    May 24, 2018 at 10:19

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