Knowlpedia

Lemma

Absorbing logarithmic losses into a power

Every fixed power of a logarithm grows more slowly than any negative power of a small parameter.

Statement

For fixed a>0a>0 and m≥0m\ge0,

εa(1+∣log⁡ε∣)m⟶0(ε↓0).\varepsilon^a(1+|\log\varepsilon|)^m\longrightarrow0 \qquad(\varepsilon\downarrow0).

Consequently, for b<ab<a,

εa(1+∣log⁡ε∣)m=o(εb).\varepsilon^a(1+|\log\varepsilon|)^m=o(\varepsilon^b).

The factor can be absorbed by spending an arbitrarily small positive amount of power.

Proof and uniformity

Put s=−log⁡εs=-\log\varepsilon. The expression becomes e−as(1+s)me^{-as}(1+s)^m. Choose an integer n>mn>m and use eas/2≥(as/2)n/n!e^{as/2}\ge(as/2)^n/n!; the resulting bound tends to zero. Constants depend on a,ma,m. The conclusion does not give a common threshold for unbounded mm without extra information.