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In order to ﬁnish the proof from last time we need a lemma.

Lemma 1.1

Let φ be a positive and increasing function on the positive reals, and let α, c be positive constants.

For all 0 < γ < α there is � > 0 such that φ ( r ) ≤ c �� r � α s � + � φ ( s ) (1) for 0 < r < s implies φ ( r ) ≤ c � � r � γ s φ ( s ) (2) for some constant c � .

Proof Choose 0 < τ < 1 such that � ≤ τ α .

Then φ ( τ s ) ≤ c ( τ α + � ) φ ( s ) ≤ 2 cτ α φ ( s ) .

Therefore φ ( τ k s ) ≤ (2 cτ α ) k φ ( s ) .

Pick γ so that 2 cτ α − γ ≤ 1 and we have (3) (4) φ ( τ k s ) ≤ τ kγ φ ( s ) .

(5) When r = τ k s this is precisely what we wanted with c � = 1.

If instead τ k +1 s then ≤ r ≤ τ k s φ ( r ) ≤ φ ( τ k s ) ≤ τ kγ φ ( s ) ≤ 1 � r � γ φ ( s ) τ s (6) which is what we needed.

Note that by taking τ very small we can get γ as close to α we like.

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Now we apply this to what we were doing last time.

Let φ ( r ) = � B r ( x 0 ) |� u | 2 .

We showed that φ ( r ) ≤ �� r � n s + k || A ij − A ij ( x 0 ) || � φ ( s ) (7) our By picking lemma.

s Pick we can get || A ij − A ij ( x 0 ) || as small as we like so we will be able to apply 0 < β < 1 and set γ = n − 2 + 2 β .

By our lemma there is a constant k � with φ ( r ) ≤ k � � r � n − 2+2 β φ ( s ) .

s (8) In our old notation this is � B r ( x 0 ) |� u | ≤ k � � r � n − 2+2 β � s B s ( x 0 ) |� u | 2 , (9) and Holder continuity follows by Morrey’s lemma.

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