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parse/train/_WnwtieRHxM/_WnwtieRHxM.md

@@ -352,7 +352,7 @@ $$
 \begin{array} { r l } & { \bullet \operatorname* { l i m } _ { t  \infty } \sum _ { i = 1 } ^ { t } \| \nabla L _ { \lambda } ( \pmb { \theta } ^ { ( t ) } ; \mathbf { w } ) \| < \infty ; } \\ & { \bullet \operatorname* { l i m } _ { t  \infty } \nabla L _ { \lambda } ( \pmb { \theta } ^ { ( t ) } ; \mathbf { w } ) = 0 . } \end{array}
 $$
 
-Now we need to show that under appropriate learning rate, which is specified in Lemma A.1, gradient descent converges to the stationary point that corresponds to the zero risk under weak regularization. Using the result from Lemma A.1, notice that if $L _ { \lambda } ( \pmb \theta ^ { ( t ) } ; \mathbf { w } )$ does not decrease to 0, then the denominator Lλ(θ(t); w)2 log 1L (θ(t);w) is bounded from below.
+Now we need to show that under appropriate learning rate, which is specified in Lemma A.1, gradient descent converges to the stationary point that corresponds to the zero risk under weak regularization. Using the result from Lemma A.1, notice that if $L _ { \lambda } ( \pmb \theta ^ { ( t ) } ; \mathbf { w } )$ does not decrease to 0, then the denominator Lλ(θ(t); w)2 log 1L (θ(t);w) is bounded from below.
 
 However, there exists a constant learning rate such that $\textstyle \sum _ { i = t _ { 0 } } ^ { t } \eta _ { i } \to \infty$ as $t \to \infty$ , which leads to contradiction. Therefore, for weighted ERM with weak regularization, gradient descent converges to the stationary point where $L _ { \lambda } ( \pmb \theta ^ { ( \bar { t } ) } ; \mathbf { w } ) = 0$ .