Let $$ Gamma (f, g): = frac12f & g; g; ; ; text {for} f, g in C ^ 1 ( mathbb R), $$ $ mu $ to be a measure of probability on $ ( mathbb R, mathcal B ( mathbb R)) $ with a continuously differentiable and positive density $ varrho $ regarding the Lebesgue measure $ lambda $, $$ mathcal E (f, g): = int Gamma (f, g) : { rmd} mu $$ and $$ Af: = frac12f & # 39; + frac12 ( ln varrho) & # 39; f & # 39; ; ; ; text {for} f in C ^ 2 ( mathbb R). $$ (For the simplicity of the notation, $ Gamma (f): = Gamma (f, g) $ and $ mathcal E (f): = mathcal E (f, f) $.)

Now let $ v in C ^ 2 ( mathbb R) $ with $ v ge1 $. Suppose we know that $$ – mu left ( frac {Av} v | g | ^ 2 right) the math E (g) tag1 ; ; ; text {for all} g in C_c ^ infty ( mathbb R). $$ In addition, suppose there is a $ ( zeta_k) _ {k in mathbb N} subseteq C_c ^ infty ( mathbb R) $ with $$ 0 the zeta_k the zeta_ {k + 1} ; ; ; text {for all} k in mathbb N tag2 $$ and $$ zeta_k xrightarrow {k to infty} 1 ; ; ; mu text {- almost surely} tag3 $$ as good as $$ Gamma ( zeta_k) the frac1k ; ; ; text {for all} k in mathbb N. tag4 $$

Now let $ f in C_c ^ infty ( mathbb R) $ and $ m in mathbb R $. How can we show $ (1) $ for $ g: = f-m $?

The idea is that $ g_k: = zeta_kg $ belongs to $ C_c ^ infty ( mathbb R) $ (contrary to $ g $) and so $$ – mu left ( frac {Av} v | g_k | ^ 2 right) the math E (g_k) tag5 ; ; ; text {for all} g in C_c ^ infty ( mathbb R) ; ; ; text {for all} k in mathbb N. $$ Now, $$ | Gamma (g_k) | le | Gamma (g) | +2 | g | sqrt { Gamma (g)} + g ^ 2 tag6 $$ and $$ | Gamma (g_k) – Gamma (g) | le | zeta_k ^ 2-1 || Gamma (g) | + frac2 { sqrt k} | g_k | sqrt { Gamma (g)} + frac1kg ^ 2 xrightarrow {k to infty} 0 tag7 $$ and so $$ mathcal E (g_k) xrightarrow {k to infty} mathcal E (g) tag8 $$ by the dominated convergence theorem.

The left side of $ (5) $ seems to be more complicated. Clearly, $ frac {Av} v | g_k | ^ 2 xrightarrow {k to infty} frac {Av} v | g | ^ 2 $but $ frac {Av} v | g_k | ^ 2 $ does not seem to be dominated by an integrable function.