Let T : X → Y be a linear map

1. Derivatives Let (X, k · k_X) and (Y, k · k_Y) be normed vector spaces.

Definition 1.1. A linear map T : X → Y is said to be bounded if there exists M > 0 such that

kT (x)k_Y ≤ M kxk_X for any x ∈ X.

Proposition 1.1. Let T : X → Y be a linear map. The following three statements are equivalent.

(1) T is bounded.

(2) T is continuous.

(3) T is continuous at 0.

Proof. The proof is left to the reader as an exercise. 

Definition 1.2. Let f : U → Y be a function defined on a nonempty open subset U of X.

We say that f is differentiable at x₀ ∈ U if there exists a bounded linear map T : X → Y such that

khklim_X→0

kf (x₀+ h) − f (x₀) − T (h)k_Y

khk_X = 0.

In this case, T is called a derivative of f at x0.

Proposition 1.2. If f is differentiable at x0, its derivative is unique.

The proof follows from the following fact:

Lemma 1.1. Let S : X → Y be a linear map such that S(x) = 0 for any x ∈ X with kxk_X = 1. Then S = 0 is the zero linear map.

Proof. Since S is linear, S(0X) = 0Y. Here 0X and 0Y are the zero vectors of X and of Y respectively. For x 6= 0_X, set w = x/kxk_X. Then w is a unit vector and x = kxk_Xw. Since w is a unit vector, S(w) = 0Y. We see that

S(x) = S(kxk_Xw) = kxk_XS(w) = kxk_X · 0_Y = 0_Y.

This shows that S = 0 is the zero linear map. 

Let us go back to prove Proposition 1.2. Since f is differentiable at x0, there exist bounded linear maps T₁, T₂: X → Y such that for any  > 0, there exists δ > 0 with

kf (x₀+ h) − f (x0) − T1(h)kY ≤  2khk_X, kf (x₀+ h) − f (x₀) − T₂(h)k_Y ≤ 

2khk_X,

whenever 0 < khk_X < δ. By triangle inequality for norms, we obtain that whenever 0 <

khk_X < δ,

kT₁(h) − T2(h)k ≤ kf (x0+ h) − f (x0) − T1(h)kY + kf (x0+ h) − f (x0) − T2(h)kY

≤ 

2khk_X+ ≤ 

2khk_X = khkX.

Let S = T1 − T₂ and xh = h/khkX for 0 < khkX < δ. Then xh is a unit vector and kS(x_h)k_Y <  for any  > 0. This implies that kS(x)k = 0 for any x ∈ X with kxk_X = 1.

Lemma 1.1 implies that S = 0 and hence T₁ = T₂.

Remark. Since the derivative of f at x₀ is unique if it is differentiable at x₀, its derivative is denoted by Df (x0).

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Example 1.1. Let f : R² → R², f (x, y) = (x²− y², 2xy). Show that f is differentiable at p(2, 1) and find Df (2, 1).

Proof. Let h = (h1, h2) ∈ R². Then

f (p + h) − f (p) = (4h1− 2h₂+ h²₁− h²₂, 2h1+ 4h2+ 2h1h2).

Take T (h) = (4h1− 2h₂, 2h1+ 4h2). Then T is a linear map and f (p + h) − f (p) − T (h) = (h²₁− h²₂, 2h1h2).

We find that

kf (p + h) − f (p) − T (h)k = q

(h²₁− h²₂)²+ (2h₁h₂)² = khk² and that

khk→0lim

kf (p + h) − f (p) − T (h)k

khk = lim

khk→0khk = 0.

Since any linear map T on R² is bounded, the above equation implies that f is differentiable at p and Df (2, 1) = T.

 Proposition 1.3. If f is differentiable at x₀, then it is continuous at x₀.

Proof. Since f is differentiable at x0, choose δ1> 0 such that kf (x₀+ h) − f (x0) − T (h)kY < khkX

whenever 0 < khk_X < δ1. Here T = Df (x0). Since T is bounded, there exists M > 0 such that kT (x)k_Y ≤ M kxk_X. For any x ∈ B(x0, δ₁), we have kx − x₀k < δ₁. Thus

kf (x) − f (x₀)kY ≤ kf (x) − f (x₀) − T (x − x0)kY + kT (x − x0)kY

≤ kx − x₀k_X + M kx − x0k_x

= (M + 1)kx − x₀k_X. Let C = M + 1. We have

kf (x) − f (x₀)k_Y ≤ Ckx − x₀k_X

whenever x ∈ B(x0, δ). This implies that f is continuous at x0.  Example 1.2. Let X = C([0, 1]) be the space of all real valued continuous functions on [0, 1] equipped with the sup norm

kf k_∞= sup

x∈[0,1]

|f (x)|

and Y = R be the space of all real numbers with absolute value. Define F : X → Y by F (u) =

Z 1 0

u(x)²dx.

Show that F is differentiable at all u₀∈ X.

Proof. We observe that

F (u₀+ h) − F (u₀) = Z 1

0

(2u₀(x)h(x) + h(x)²)dx for any h ∈ X. Define T : X → R by

T (h) = 2 Z 1

0

u0(x)h(x)dx, h ∈ X.

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For any h ∈ X,

|F (u₀+ h) − F (u0) − T (h)| =    

Z 1 0

h(x)²dx    

≤ khk²_∞ and hence

0 ≤ |F (u₀+ h) − F (u0) − T (h)|

khk_∞ ≤ khk∞.

By Sandwich principle,

khklim∞→0

|F (u₀+ h) − F (u0) − T (h)|

khk_∞ = 0.

Hence if we can show that T is a bounded linear map, then F is differentiable at u₀ with DF (u₀) = T.

We leave it to the reader to verify that T is a linear map. Now let us prove that T is bounded. If u₀ = 0, T is the zero linear map and hence T is bounded. Assume that u₀ 6= 0.

For any h ∈ X, we have

|T (h)| =    

Z 1 0

2u₀(x)h(x)dx    

≤ 2ku₀k∞khk∞.

This implies that T is bounded. We complete the proof of our assertion.