Multivariable Advanced Calculus

122 THE DERIVATIVEor equivalently,||f (y) − f (x) − Df (x) (y − x)||lim= 0. (6.3)y→x||y − x||The symbol, o (v) should be thought of as an adjective. Thus, if t and k are constants,o (v) = o (v) + o (v) , o (tv) = o (v) , ko (v) = o (v)and other similar observations hold.Theorem 6.1.2 The derivative is well defined.Proof: First note that for a fixed vector, v, o (tv) = o (t). This is becauseo (tv)lim = lim ||v|| o (tv) = 0t→0 |t| t→0 ||tv||Now suppose both L 1 and L 2 work in the above definition. Then let v be any vectorand let t be a real scalar which is chosen small enough that tv + x ∈ U. Thenf (x + tv) = f (x) + L 1 tv + o (tv) , f (x + tv) = f (x) + L 2 tv + o (tv) .Therefore, subtracting these two yields (L 2 − L 1 ) (tv) = o (tv) = o (t). Therefore, dividingby t yields (L 2 − L 1 ) (v) = o(t)t. Now let t → 0 to conclude that (L 2 − L 1 ) (v) =0. Since this is true for all v, it follows L 2 = L 1 . This proves the theorem. Lemma 6.1.3 Let f be differentiable at x. Then f is continuous at x and in fact,there exists K > 0 such that whenever ||v|| is small enough,Also if f is differentiable at x, then||f (x + v) − f (x)|| ≤ K ||v||o (∥f (x + v) − f (x)∥) = o (v)Proof: From the definition of the derivative,f (x + v) − f (x) = Df (x) v + o (v) .Let ||v|| be small enough that o(||v||)||v||< 1 so that ||o (v)|| ≤ ||v||. Then for such v,||f (x + v) − f (x)|| ≤ ||Df (x) v|| + ||v||≤(||Df (x)|| + 1) ||v||This proves the lemma with K = ||Df (x)|| + 1.The last assertion is implied by the first as follows. Define{o(∥f(x+v)−f(x)∥)h (v) ≡ ∥f(x+v)−f(x)∥if ∥f (x + v) − f (x)∥ ̸= 00 if ∥f (x + v) − f (x)∥ = 0Then lim ∥v∥→0 h (v) = 0 from continuity of f at x which is implied by the first part.Also from the above estimate,o (∥f (x + v) − f (x)∥)∥f (x + v) − f (x)∥∥ ∥v∥ ∥ = ∥h (v)∥ ≤ ∥h (v)∥ (||Df (x)|| + 1)∥v∥This establishes the second claim. Here ||Df (x)|| is the operator norm of the linear transformation, Df (x).