f0(x0)h for any h ∈ R

Let X and Y be normed vector spaces¹ and U be an open subsets of X. We say that a function f : U → Y is differentiable at x0 if there exists a bounded linear map T : X → Y such that

khklim_X→0

|f (x₀+ h) − f (x0) − T (h)kY

khk_X = 0.

We denote T by Df (x₀). If f is differentiable at every point of U, we say that f is differen- tiable on U.

Example 1.1. If f : (a, b) → R is differentiable at x0, then Df (x0)(h) = f⁰(x0)h for any h ∈ R.

Suppose that f is differentiable at every points of U. Then Df defines a map Df : U → L^b(X, Y ), x 7→ Df

where L^b(X, Y ) is the space of bounded linear maps from X to Y. We equip L^b(X, Y ) with the operator norm defined by

kT k = max

khk_X=1

kT (h)k_Y for T ∈ L(X, Y ).

Then L^b(X, Y ) is a normed space. Since L^b(X, Y ) is a normed space, we can discuss the continuities and differentiabilities of Df. If Df : U → Y is continuous on U, we say that f is continuously differentiable on U. The set of all continuously differentiable functions from U to Y is denoted by C¹(U, Y ). We denote C¹(U, R) (or C¹(U, C)) by C¹(U ).

Definition 1.1. We say that a function f : U → Y is twice differentiable at x0 if there exists δ > 0 such that Df (x) exists at every point of x ∈ B(x0, δ) and the function

Df : B(x0, δ) → L^b(X, Y )

is differentiable at x0. We denote D(Df )(x0) by D²f (x0). If f is twice differentiable at every point of U, we say that f is twice differentiable on U.

Let us look at the example when X = Y = Rⁿ. Suppose that f : (a, b) → R is twice differentiable at x₀, i.e. Df : (a, b) → L^b(R, R) is differentiable at x0. We remark that L^b(R, R) coincides with L(R, R). Then there exists a linear map T : R → L(R, R) such that

k→0lim

kDf (x₀+ k) − Df (x0) − T (k)k_L(R,R)

|k| = 0.

For  > 0, there exists δ > 0 so that

kDf (x₀+ k) − Df (x0) − T (k)k_L(R,R) < |k|

whenever 0 < |h| < δ. By the norm inequality,

|Df (x₀+ k)(h) − Df (x0)(h) − T (k)(h)| ≤ kDf (x0+ k) − Df (x0) − T (k)k_L(R,R)|h| < |h||k|.

Since Df (x)(h) = f⁰(x)h for any x ∈ (a, b), we have

Df (x₀+ k)(h) − Df (x₀)(h) = (f⁰(x₀+ k) − f⁰(x₀))h.

Dividing the above inequality by k, we obtain 

f⁰(x0+ k) − f⁰(x0)

k h − T (1) (h)    

< |h|.

1All the normed vector spaces in this note are assumed to be real or complex.

1

T (1)(h) = lim

k→0

f⁰(x0+ k) − f⁰(x0)

k h = f⁰⁰(x₀)h.

By linearity of T, we obtain that T (k)(h) = f⁰⁰(x0)hk. In other words, (D²f )(x₀)(k)(h) = f⁰⁰(x₀)hk.

Let us recall the basic definition of bilinear maps. Let X, Y, Z be normed vector spaces.

A bilinear map T : X × Y → Z is a function such that (1) T (ax1+ bx2, y) = aT (x1, y) + bT (x2, y)

(2) T (x, ay1+ by2) = aT (x, y1) + bT (x, y2)

for any a, b ∈ R (or C) and for any x, x1, x₂ ∈ X and for any y, y₁, y₂ ∈ Y. We say that T is a bounded bilinear map if there exists M > 0 so that kT (x, y)kZ ≤ M kxk_Xkyk_Y. The space of bounded bilinear maps from X × Y to Z is denoted by bil(X × Y, Z). In general, we consider the n-linear maps

Definition 1.2. Let X₁, · · · , X_nand Y be normed vector spaces. We can define the notion of n-linear maps

T : X1× · · · × X_n→ Y.

We say that T is bounded if there exists M > 0 so that kT (x₁, · · · , x_n)k_Y ≤ M

n

Y

i=1

kx_ik_Xi for any (x₁, · · · , x_n) ∈ X₁× · · · × X_n.

The space of all bounded n-linear maps from X1× · · · × X_n to Y is denoted by mult(X1× · · · × X_n, Y ) = {T : X1× · · · × X_n→ Y : T is bounded n-linear}.

The norm of a bounded n-linear map T is defined to be kT k = sup

kxik_Xi=1, 1≤i≤n

kT (x₁, · · · , x_n)k_Y. Then we have the following results:

Proposition 1.1. mult(X1× · · · × X_n, Y ) is a normed vector space. Furthermore, if Y is a Banach space, so is mult(X₁× · · · × X_n, Y ).

Proof. This is left to the reader as an exercise. 

Proposition 1.2. Let X, Y, Z be normed vector spaces. We have the following isomorphism of normed vector spaces

L^b(X, L^b(Y, Z)) ∼= bil(X × Y, Z).

Proof. Suppose T : X → L^b(Y, Z) is a linear map. For each x ∈ X, T (x) ∈ L^b(Y, Z), i.e.

T (x) : Y → Z is a bounded linear map for any x ∈ X. We define ϕT : X × Y → Z by ϕ_T(x, y) = T (x)(y). Then ϕ_T is bilinear. Now let us prove that ϕ_T is bounded. Since T is bounded, we can find M > 0 so that kT (x)k_L^b_(Y,Z) ≤ M kxk_X for any x ∈ X. Since T (x) ∈ L^b(Y, Z), kT (x)(y)k_Z ≤ kT (x)k_Lb(Y,Z)kyk_Y for any y ∈ Y. For any x ∈ X and Y ∈ Y, one has

kϕ_T(x, y)kZ= kT (x)(y)kZ ≤ kT (x)k_Lb(Y,Z)kyk_Y ≤ M kxk_Xkyk_Y.

This shows that ϕT ∈ bil(X × Y, Z). We obtain a map

ϕ : L^b(X, L^b(Y, Z)) → bil(X × Y, Z), T 7→ ϕ_T.

We leave it to the reader to check that ϕ is an isomorphism of normed vector spaces. We

complete the proof of our assertion. 

If f : U → Y is twice differentiable on U, Df : U → L^b(X, Y ) is differentiable on U.

Therefore D²f defines a map from U into L^b(X, L^b(X, Y )). By Proposition 1.2, L^b(X, L^b(X, Y )) ∼= bil(X × X, Y ).

Hence we obtain a map

D²f : U → bil(X × X, Y ).

If D²f is continuous, we say that f : U → Y is twice continuously differentiable on U. The set of all twice continuously differentiable functions from U into Y is denoted by C²(U, Y ).

The set C²(U, R) (or C²(U, C)) is denoted by C²(U ) for simplicity.

Inductively, we say that f is j-times differentiable on U if D^j−1f exists at every point of U and the map D^j−1f : U → multj−1(X^j−1, Y ) is differentiable on U. We denote D^jf (x) = D(D^j−1f )(x) for x ∈ U when f is j-times differentiable on U. If f is k-times differentiable on U, we obtain a function

D^jf : U → mult_j(X^j, Y ).

When D^jf is continuous on U, we say that f is j-times continuously differentiable on U. The set of all j-times continuously differentiable functions from U to Y is denoted by C^j(U, Y ).

The set C^j(U, R) (or C^j(U, C)) is simply denoted by C^j(U ).

Definition 1.3. Let U be an open subset of a normed space X. We set C^∞(U ) =

∞

\

j=0

C^j(U ).

We call C^∞(U ) the space of real valued smooth functions on U ; elements of C^j(U ) are called smooth functions on U.

We can check the following results.

Example 1.2. Let f ∈ C^j(a, b). For each 1 ≤ i ≤ j, the map (Dⁱf )(x) : (a, b) → mult_i(Rⁱ, R)

is given by (Dⁱf )(x)(h₁, · · · , h_i) = f⁽ⁱ⁾(x)h₁· · · h_i at any point x ∈ (a, b). Furthermore, Dⁱf (x)(h, · · · , h) = fⁱ(x)hⁱ for any h ∈ R. The Taylor Theorem can be rewritten as

f (x + h) =

j−1

X

i=0

(Dⁱf )(x)(h, · · · , h)

i! +D^jf (x + ch)(h, · · · , h) j!

where h is a real number so that |h| < δ for some δ > 0 and c ∈ [0, 1].

Example 1.3. Let U be an open subset of Rⁿ and f : U → R be a function. Suppose that f ∈ C^j(U ) when j is sufficiently large (or we may assume that f is smooth). Let us compute D²f : U → L(Rⁿ× Rⁿ, R). Let h = (h1, · · · , h_n) and k = (k₁, · · · , k_n). Then for x ∈ U,

(1.1) (D²f (x))(h, k) =

n

X

i,j=1

∂²f

∂x_i∂x_j(x)hikj.

f : U → L(R × R × R , R). Assume l = (l (D³f (x))(h, k, l) =

n

X

i,j,s=1

∂³f

∂xi∂xj∂xs

(x)h_ik_jl_s.

Inductively it is possible for us to write down all the formulas for Dⁱf (x) for any i ≥ 0.

Recall that

H_i(f )(x)(h) = X

|α|=i

 i α



(D^αf )(x)h^α.

It is easy for us to see that H_i(f )(x)(h) = (Dⁱf )(x)(h, · · · , h) for any h ∈ Rⁿ. Let us prove (1.1). Let T = D²f (x). For any  > 0, there exists δ > 0 such that

kDf (x + k) − Df (x) − T (k)k_L(Rn,R)< kkk whenever 0 < kkk < δ. By norm inequality,

|Df (x + k)(h) − Df (x)(h) − T (k)(h)| ≤ kDf (x + k) − Df (x) − T (k)k_L(Rn,R)khk < khkkkk.

Choose h = e_i where {e₁, · · · , e_n} is the standard basis for Rⁿ. We know that Df (x)(ei) = ∂f

∂e_i(x).

We find that

∂f

∂x_i(x + k) − ∂f

∂x_i(x) − T (k)(e_i)    

< kkk.

Let k = tej; we find that when 0 < |t| < δ, 

∂f

∂xi

(x + te_j) − ∂f

∂xi

(x) − T (h)(e_i)    

< |t|.

Dividing the above equation by |t|, we obtain that 

f_xi(x + te_j) − f_xi(x)

t − T (e_j)(e_i)    

<  whenever 0 < |t| < δ. This shows that

fxixj(x) = lim

t→0

fxi(x + tej) − fxi(x)

t = T (ej)(ei).

If we write h =Pn

i=1hiei and k =Pn

j=1kjej, then T (k)(h) =

n

X

i,j=1

h_ik_jT (e_j)(e_i) =

n

X

i,j=1

f_xixj(x)h_ik_j. This shows that

D²f (x)(h, k) = T (k)(h) =

n

X

i,j=1

f_xixj(x)h_ik_j. Inductively, we are able to determine all D^jf (x) for any j ≥ 1.

Example 1.4. Let f (x, y) = e^x+2y. Compute Df (0, 0), D²f (0, 0) and D³f (0.0).

Answer: let h = (h1, h2) and k = (k1, k2) and l = (l1, l2) in R² Then Df (x, y)(h) = f_x(x, y)h₁+ f_x(x, y)h₂

D²f (x, y)(h, k) = f_xx(x, y)h₁k₁+ f_xy(x, y)h₁k₂+ f_yx(x, y)h₂k₁+ f_yy(x, y)h₂k₂ D³f (x, y)(h, k, l) = fxxx(x, y)h1k1l1+ hyyy(x, y)h2k2l2

+ f_xxy(x, y)h₁k₁l₂+ f_xyx(x, y)h₁k₂l₁+ f_yxx(x, y, z)h₂k₁l₁ + f_xyy(x, y)h₁k₂l₂+ f_yxy(x, y)h₂k₁l₂+ f_yyx(x, y)h₂k₂l₁.

Remark. Let U be an open subset of Rⁿ. We remark that f ∈ C^j(U ) if and only if D^αf exists and continuous on U for any n-dimensional multiindices α with |α| = j. It follows from the fact that any norms on finite dimensional (real) vector spaces are equivalent.

Department of Mathematics, National Cheng Kung University, Taiwan, fjmliou@mail.ncku.edu.tw NCTS, Mathematics