(* This does some calculus for Lemma 2.1 and Lemma 2.4 *) 
(* For Lemma 2.1 *) 

f1[v_]:= 1/2 (Log[1+ 2v^2]) + Log[1 + Sqrt[2] v]

Series[f1[v], {v, 0, 3}]

Plot[f1[x], {x, 0, 3}]

Simplify[D[f1[v], v]]

(((Sqrt[2])/(1+(Sqrt[2]) v))+((2 v)/(1+2 (v^2))))/.{v-> 0}

(* We take the 2nd derivatives and find its maximum*)

D[D[f1[v], v], v]

Plot[-(2/(((1+(Sqrt[2]) v))^2))-((8 (v^2))/(((1+2 (v^2)))^2))+(2/(1+2 (v^2))), {v, 0, 1/8}]

Together[D[-(2/(((1+(Sqrt[2]) v))^2))-((8 (v^2))/(((1+2 (v^2)))^2))+(2/(1+2 (v^2))), v]]

Collect[4 ((Sqrt[2])-6 v-12 (Sqrt[2]) (v^2)-32 (v^3)+12 (Sqrt[2]) (v^4)+24 (v^5)+16 (Sqrt[2]) (v^6)), v]

(* We show that the second derivative is a strictly increasing function *)

FindMinimum[{((4 ((Sqrt[2])-6 v-12 (Sqrt[2]) (v^2)-32 (v^3)+12 (Sqrt[2]) (v^4)+24 (v^5)+16 (Sqrt[2]) (v^6)))/((((1+(Sqrt[2]) v))^3) (((1+2 (v^2)))^3))), 0 < v < 1/8},{v, 1/16}]

Plot[-(2/(((1+(Sqrt[2]) v))^2))-((8 (v^2))/(((1+2 (v^2)))^2))+(2/(1+2 (v^2))), {v, 0, 1/8}]

(* The maximum of the second derivative *)

N[(-(2/(((1+(Sqrt[2]) v))^2))-((8 (v^2))/(((1+2 (v^2)))^2))+(2/(1+2 (v^2))))/.{v-> 1/8}]

N[Sqrt[2] + 1/2*(0.377608)*(1/8)]

(* ArcTan Lemma 2.3 *)

Plot[ t*(Pi/2 - ArcTan[t/2]), {t, 0, 10}, PlotRange->{{0, 10},{0,2.2}}]

Simplify[D[t*(Pi/2 - ArcTan[t/2]), t]]

Together[D[Pi - 2ArcTan[t/2], t]-D[4t/(4+t^2), t]]

Both Pi - 2ArcTan[t/2] and 4t/(4+t^2) goes to zero as t -> infty, and The difference of the differenital is negative. This means that the first function Pi- 2ArcTan[2/2] is bigger. Hence, the  (1/2) (π-((4 t)/(4+(t^2)))-2 ArcTan[(t/2)]) is positive. Hence, t*(Pi/2 - ArcTan[t/2]) is strictly increasing.

Plot[(1/2) (π-((4 t)/(4+(t^2)))-2 ArcTan[(t/2)]), {t, 0, 10},PlotRange->{{0, 10},{0,2.2}}]