HW-1: Spivak Problems

Part 1: Inner-product preserving \(\implies\) Norm preserving

Assume \(T\) preserves inner products. For any \(x, y \in \mathbf{R}^n\): \[ \langle Tx, Ty \rangle = \langle x, y \rangle \]

Let \(y = x\). Substituting this gives: \[ \langle Tx, Tx \rangle = \langle x, x \rangle \]

Using the standard relation between a vector's norm and inner product (\(\Vert x \Vert^2 = \langle x, x \rangle\)): \[ \Vert Tx \Vert^2 = \Vert x \Vert^2 \]

Since norms are non-negative, taking the square root yields: \[ \Vert Tx \Vert = \Vert x \Vert \]

Thus, \(T\) is norm preserving.

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Part 2: Norm preserving \(\implies\) Inner-product preserving

Assume \(T\) preserves norms, meaning \(\Vert Tx \Vert = \Vert x \Vert\) for all \(x \in \mathbf{R}^n\).

Recall the polarization identity for any vectors \(x\) and \(y\): \[ \langle x, y \rangle = \frac{1}{2} (\Vert x+y \Vert^2 - \Vert x \Vert^2 - \Vert y \Vert^2) \]

Apply this identity to vectors \(Tx\) and \(Ty\): \[ \langle Tx, Ty \rangle = \frac{1}{2} (\Vert Tx+Ty \Vert^2 - \Vert Tx \Vert^2 - \Vert Ty \Vert^2) \]

Because \(T\) is a linear transformation, \(Tx + Ty = T(x+y)\): \[ \langle Tx, Ty \rangle = \frac{1}{2} (\Vert T(x+y) \Vert^2 - \Vert Tx \Vert^2 - \Vert Ty \Vert^2) \]

By our assumption, \(T\) preserves the norm of any vector (\(\Vert Tv \Vert = \Vert v \Vert\)). Applying this gives: \[ \langle Tx, Ty \rangle = \frac{1}{2} (\Vert x+y \Vert^2 - \Vert x \Vert^2 - \Vert y \Vert^2) \]

The right side is simply the polarization identity for \(\langle x, y \rangle\). Therefore: \[ \langle Tx, Ty \rangle = \langle x, y \rangle \]

Thus, \(T\) is inner-product preserving.

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Part 3: Prove \(T\) is 1-1 (Injective)

A linear transformation is injective if its kernel contains only the zero vector (\(\ker(T) = \{0\}\)).

Assume \(Tx = 0\). Taking the norm of both sides: \[ \Vert Tx \Vert = 0 \]

Since \(T\) is norm preserving, \(\Vert Tx \Vert = \Vert x \Vert\). Therefore: \[ \Vert x \Vert = 0 \]

A vector's norm is zero if and only if the vector is zero, so \(x = 0\).

This proves \(\ker(T) = \{0\}\), meaning \(T\) is 1-1.

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Part 4: Prove \(T^{-1}\) is of the same sort

Since \(T: \mathbf{R}^n \rightarrow \mathbf{R}^n\) is a 1-1 linear transformation mapping a finite-dimensional space to itself, standard linear algebra dictates that \(T\) is also onto (surjective). Therefore, the inverse \(T^{-1}\) exists.

Let \(y \in \mathbf{R}^n\). Because \(T\) is surjective, there is a unique vector \(x \in \mathbf{R}^n\) such that: \[ Tx = y \implies x = T^{-1}y \]

Taking the norm of the first equation gives: \[ \Vert Tx \Vert = \Vert y \Vert \]

Since \(T\) is norm preserving, replace \(\Vert Tx \Vert\) with \(\Vert x \Vert\): \[ \Vert x \Vert = \Vert y \Vert \]

Finally, substitute \(x = T^{-1}y\) back into the equation: \[ \Vert T^{-1}y \Vert = \Vert y \Vert \]

This proves that \(T^{-1}\) is also norm preserving (and by Part 1, inner-product preserving).

By the standard relationship between a vector's norm and its inner product, \(\Vert v \Vert^2 = \langle v, v \rangle\).

Starting with the left-hand side, we expand the squared norm: \[ \Vert x + y \Vert^2 = \langle x + y, x + y \rangle \]

Using the bilinearity (distributive property) of the inner product: \[ \langle x + y, x + y \rangle = \langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle \]

Since the inner product over a real vector space is symmetric, \(\langle x, y \rangle = \langle y, x \rangle\): \[ = \langle x, x \rangle + 2\langle x, y \rangle + \langle y, y \rangle \]

Because \(x\) and \(y\) are given as orthogonal, their inner product is zero (\(\langle x, y \rangle = 0\)). Substituting this into the equation: \[ = \langle x, x \rangle + 2(0) + \langle y, y \rangle \]

\[ = \langle x, x \rangle + \langle y, y \rangle \]

Finally, applying the definition of the norm (\(\langle x, x \rangle = \Vert x \Vert^2\) and \(\langle y, y \rangle = \Vert y \Vert^2\)): \[ = \Vert x \Vert^2 + \Vert y \Vert^2 \]

Therefore, \(\Vert x + y \Vert^2 = \Vert x \Vert^2 + \Vert y \Vert^2\). ∎

\(A\) is closed \(\implies\) Its complement \(\mathbf{R}^n \setminus A\) is an open set.

\(x \notin A \implies x \in \mathbf{R}^n \setminus A\).

By the definition of an open set, there exists a radius \(d > 0\) such that the open ball centered at \(x\) is entirely contained in the complement: \[ B(x, d) = \{z \in \mathbf{R}^n : \vert z - x \vert < d\} \subset \mathbf{R}^n \setminus A \]

This implies \(B(x, d) \cap A = \emptyset\).

Therefore, no point \(y \in A\) lies inside \(B(x, d)\): \[ \vert y - x \vert \ge d \quad \forall y \in A \] ∎

Let \(b \in B\). Since \(A \cap B = \emptyset \implies b \notin A\).

By Part (a), \(\exists d_b > 0\) such that: \[ \vert y - b \vert \ge d_b \quad \forall y \in A \]

Define an open ball around each \(b \in B\) with radius \(\frac{d_b}{2}\): \(U_b = B\left(b, \frac{d_b}{2}\right)\).

For any \(x \in U_b\), applying the triangle inequality for any \(y \in A\): \[ \vert y - b \vert \le \vert y - x \vert + \vert x - b \vert \implies \vert y - x \vert \ge \vert y - b \vert - \vert x - b \vert \]

Since \(\vert y - b \vert \ge d_b\) and \(\vert x - b \vert < \frac{d_b}{2}\): \[ \vert y - x \vert > d_b - \frac{d_b}{2} = \frac{d_b}{2} \]

The collection of open balls \(\{U_b\}_{b \in B}\) forms an open cover for \(B\).

Since \(B\) is compact, this open cover has a finite subcover. Thus, there exist \(b_1, b_2, \dots, b_k\) such that: \[ B \subset U_{b_1} \cup U_{b_2} \cup \dots \cup U_{b_k} \]

Let \(d\) be the minimum of these finite radii: \[ d = \min\left(\frac{d_{b_1}}{2}, \frac{d_{b_2}}{2}, \dots, \frac{d_{b_k}}{2}\right) \]

Since the set is finite and all terms are positive, \(d > 0\).

For any \(x \in B\), \(x \in U_{b_i}\) for some \(i \in \{1, 2, \dots, k\}\).
Therefore, \(\forall y \in A\) and \(\forall x \in B\): \[ \vert y - x \vert > \frac{d_{b_i}}{2} \ge d \]

Thus, \(\vert y - x \vert \ge d\). ∎

Define two sets in \(\mathbf{R}^2\): \[ A = \{(x, y) \in \mathbf{R}^2 : y = 0\} \quad \text{(X-axis)} \] \[ B = \left\{(x, y) \in \mathbf{R}^2 : y = \frac{1}{x}, \ x \ge 1\right\} \quad \text{(Branch of a hyperbola)} \]

Verification:
Closed: Both \(A\) and \(B\) contain their boundaries/limit points, hence both are closed sets.
Not Compact: Both sets extend to infinity (\(x \to \infty\)). Since they are unbounded, neither is compact.
Disjoint (\(A \cap B = \emptyset\)): For any point in \(B\), \(y = \frac{1}{x} > 0\) (since \(x \ge 1\)). For any point in \(A\), \(y = 0\). Thus, they never intersect.

No such \(d > 0\) exists:
Consider sequences of points in \(A\) and \(B\):
\(p_n = (n, 0) \in A\) and \(q_n = \left(n, \frac{1}{n}\right) \in B\).
The distance between these points is: \[ d_n = \sqrt{(n - n)^2 + \left(\frac{1}{n} - 0\right)^2} = \frac{1}{n} \]

As \(n \to \infty\), \(d_n \to 0\).
Therefore, the distance between the two closed disjoint sets approaches \(0\), and no minimum distance \(d > 0\) exists.

Since \(U\) is an open set and \(C \subset U\), for every \(x \in C\), there exists a radius \(r_x > 0\) such that: \[ B(x, r_x) \subset U \]

Consider the open ball with half the radius, \(B\left(x, \frac{r_x}{2}\right)\). In \(\mathbf{R}^n\), its closure \(\overline{B\left(x, \frac{r_x}{2}\right)}\) is closed and bounded, making it compact (Heine-Borel Theorem).

This closed ball strictly remains inside \(U\): \[ \overline{B\left(x, \frac{r_x}{2}\right)} \subset B(x, r_x) \subset U \]

The collection of these smaller open balls forms an open cover for \(C\): \[ C \subset \bigcup_{x \in C} B\left(x, \frac{r_x}{2}\right) \]

Since \(C\) is compact, this open cover has a finite subcover. Thus, there exist finitely many points \(x_1, x_2, \dots, x_k \in C\) such that: \[ C \subset \bigcup_{i=1}^k B\left(x_i, \frac{r_{x_i}}{2}\right) \]

Define the set \(D\) as the union of the corresponding closed balls: \[ D = \bigcup_{i=1}^k \overline{B\left(x_i, \frac{r_{x_i}}{2}\right)} \]

Verification of \(D\):
\(D\) is compact: \(D\) is a finite union of compact sets, hence compact.
\(D \subset U\): Since each individual closed ball \(\overline{B\left(x_i, \frac{r_{x_i}}{2}\right)} \subset U\), their union is also contained within \(U\).
\(C \subset \text{interior}(D)\): The interior of a set contains any open sets within it. Thus: \[ \text{interior}(D) = \text{interior}\left( \bigcup_{i=1}^k \overline{B\left(x_i, \frac{r_{x_i}}{2}\right)} \right) \supset \bigcup_{i=1}^k B\left(x_i, \frac{r_{x_i}}{2}\right) \]

From the subcover defined earlier, this union of open balls contains \(C\). Therefore: \[ C \subset \text{interior}(D) \]

Thus, all conditions are satisfied: \(C \subset \text{interior}(D) \subset D \subset U\). ∎

Let \(f(x) = (f^1(x), f^2(x), \dots, f^m(x))\) and \(b = (b^1, b^2, \dots, b^m)\).

Recall the standard inequalities relating a vector's norm to its components: \[ \vert f^i(x) - b^i \vert \le \vert f(x) - b \vert \quad \forall i \in \{1, \dots, m\} \] \[ \vert f(x) - b \vert \le \sum_{i=1}^m \vert f^i(x) - b^i \vert \]

Part 1: Assume \(\lim_{x \to a} f(x) = b \implies \lim_{x \to a} f^i(x) = b^i \quad \forall i\)

Assume \(\lim_{x \to a} f(x) = b\).
By definition, for any \(\epsilon > 0\), \(\exists \delta > 0\) such that: \[ 0 < \vert x - a \vert < \delta \implies \vert f(x) - b \vert < \epsilon \]

Using the first inequality for any component \(i\): \[ \vert f^i(x) - b^i \vert \le \vert f(x) - b \vert \]

Substituting the limit condition: \[ \vert f^i(x) - b^i \vert < \epsilon \quad \forall i \]

This satisfies the precise definition of a limit for each component. Therefore: \[ \lim_{x \to a} f^i(x) = b^i \quad \forall i \]

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Part 2: Assume \(\lim_{x \to a} f^i(x) = b^i \quad \forall i \implies \lim_{x \to a} f(x) = b\)

Assume \(\lim_{x \to a} f^i(x) = b^i\) for all \(i = 1, \dots, m\).

Let \(\epsilon > 0\). By definition, for each component \(i\), \(\exists \delta_i > 0\) such that: \[ 0 < \vert x - a \vert < \delta_i \implies \vert f^i(x) - b^i \vert < \frac{\epsilon}{m} \]

Choose \(\delta = \min(\delta_1, \delta_2, \dots, \delta_m)\). Since the set is finite, \(\delta > 0\).
For \(0 < \vert x - a \vert < \delta\), the condition holds simultaneously for all \(m\) components.

Applying the second inequality: \[ \vert f(x) - b \vert \le \sum_{i=1}^m \vert f^i(x) - b^i \vert \]

Substituting the component bounds: \[ \vert f(x) - b \vert < \sum_{i=1}^m \frac{\epsilon}{m} = m \left( \frac{\epsilon}{m} \right) = \epsilon \]

Thus, for any \(\epsilon > 0\), \(\exists \delta > 0\) such that \(0 < \vert x - a \vert < \delta \implies \vert f(x) - b \vert < \epsilon\). Therefore: \[ \lim_{x \to a} f(x) = b \] ∎

Consider any straight line passing through the origin \((0,0)\).

Case 1: The vertical line (\(x = 0\))
Set \(A\) strictly requires \(x > 0\). Thus, no point on \(x = 0\) is in \(A\). The entire line is in \(\mathbf{R}^2 \setminus A\).

Case 2: Lines of the form \(y = mx\)
Subcase 2.1: \(m \le 0\)
For \(x > 0\), \(y = mx \le 0\). Since \(A\) requires \(y > 0\), the line never enters \(A\) for \(x > 0\). The entire line is in \(\mathbf{R}^2 \setminus A\).

Subcase 2.2: \(m > 0\)
A point \((x, mx)\) is in \(A\) if and only if: \[ 0 < mx < x^2 \]

Since \(x > 0\), dividing by \(x\) yields \(m < x\). This means the line only enters set \(A\) when \(x > m\). Therefore, for the interval \(x \in [-m, m]\), the points \((x, mx) \notin A\). This forms an interval around the origin that lies entirely in \(\mathbf{R}^2 \setminus A\). ∎

Step 1: \(g_h\) is continuous at \(t=0\)

From part (a), for any direction \(h\), the line \(th\) contains an interval \(t \in (-\delta, \delta)\) around \(0\) that lies entirely in \(\mathbf{R}^2 \setminus A\).

For all \(t \in (-\delta, \delta)\), \(th \notin A\), meaning: \[ g_h(t) = f(th) = 0 \]

Since \(g_h(t)\) is identically \(0\) in a neighborhood of \(t=0\), \(\lim_{t \to 0} g_h(t) = 0 = g_h(0)\). Thus, \(g_h\) is continuous at \(t=0\).

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Step 2: \(f\) is not continuous at \((0,0)\)

Since \((0,0)\) fails the condition \(x > 0\), \((0,0) \notin A\). Therefore, \(f(0,0) = 0\).

Consider the approach to the origin along the path \(y = \frac{x^2}{2}\) for \(x > 0\). For points on this path: \[ 0 < \frac{x^2}{2} < x^2 \]

These points strictly satisfy the conditions of \(A\), meaning they lie inside \(A\). Thus, \(f\left(x, \frac{x^2}{2}\right) = 1\).

Evaluating the limit as \((x,y) \to (0,0)\) along this path: \[ \lim_{x \to 0^+} f\left(x, \frac{x^2}{2}\right) = \lim_{x \to 0^+} 1 = 1 \]

Since the limit along this path (\(1\)) does not equal the function's value at the origin (\(0\)), \(f\) is not continuous at \((0,0)\). ∎

Let \(x \in \mathbf{R}^2\) be a fixed vector.

Case 1: \(x = 0\)
\(h(t) = f(0) = 0\). This is a constant function, thus differentiable (\(h'(t) = 0\)).

Case 2: \(x \neq 0\)
Evaluate \(h(t)\) for different signs of \(t\):

For \(t > 0\): \(\vert tx \vert = t\vert x \vert\) \[ h(t) = f(tx) = \vert tx \vert \cdot g\left(\frac{tx}{\vert tx \vert}\right) = t\vert x \vert \cdot g\left(\frac{x}{\vert x \vert}\right) = t \cdot f(x) \]

For \(t < 0\): \(\vert tx \vert = -t\vert x \vert\) \[ h(t) = f(tx) = -t\vert x \vert \cdot g\left(\frac{tx}{-t\vert x \vert}\right) = -t\vert x \vert \cdot g\left(-\frac{x}{\vert x \vert}\right) \]

Since \(g\) is an odd function (\(g(-y) = -g(y)\)): \[ h(t) = -t\vert x \vert \cdot \left(-g\left(\frac{x}{\vert x \vert}\right)\right) = t\vert x \vert \cdot g\left(\frac{x}{\vert x \vert}\right) = t \cdot f(x) \]

For \(t = 0\): \(h(0) = f(0) = 0 = 0 \cdot f(x)\).

Therefore, for all \(t \in \mathbf{R}\): \[ h(t) = t \cdot f(x) \]

Since \(f(x)\) is a constant real number for a fixed \(x\), \(h(t)\) is a linear function of \(t\). Linear functions are differentiable everywhere. \[ h'(t) = f(x) \] ∎

Assume \(f\) is differentiable at \((0,0)\). Let \(L = Df(0,0)\) be its derivative (a linear transformation).

Compute \(Df(0,0)\) via partial derivatives
Using the result from (a), \(f(tx) = t \cdot f(x)\):
Partial derivative with respect to \(x_1\): \[ \frac{\partial f}{\partial x_1}(0,0) = f(1,0) = \vert (1,0) \vert \cdot g(1,0) = 1 \cdot 0 = 0 \]
Partial derivative with respect to \(x_2\): \[ \frac{\partial f}{\partial x_2}(0,0) = f(0,1) = \vert (0,1) \vert \cdot g(0,1) = 1 \cdot 0 = 0 \]
Therefore, the Jacobian matrix is zero: \(L(x) = 0\).

Apply the definition of differentiability
For \(f\) to be differentiable at \((0,0)\) with \(L = 0\): \[ \lim_{x \to 0} \frac{\vert f(x) - f(0) - L(x) \vert}{\vert x \vert} = 0 \]

Substitute \(f(x)\) and \(L(x) = 0\): \[ \lim_{x \to 0} \frac{\left\vert \vert x \vert \cdot g\left(\frac{x}{\vert x \vert}\right) \right\vert}{\vert x \vert} = 0 \]

\[ \lim_{x \to 0} \left\vert g\left(\frac{x}{\vert x \vert}\right) \right\vert = 0 \]

Evaluate the limit along a specific path
Let \(x \to 0\) along a ray defined by \(x = tu\), where \(u\) is a unit vector (\(\vert u \vert=1\)) and \(t \to 0^+\). \[ \frac{x}{\vert x \vert} = \frac{tu}{t\vert u \vert} = u \]

Substitute this into the limit: \[ \lim_{t \to 0^+} \vert g(u) \vert = 0 \]

Since the expression is independent of \(t\): \[ \vert g(u) \vert = 0 \implies g(u) = 0 \]

For the limit to exist and equal \(0\), this must hold true for approaching from any direction \(u\). Therefore, \(g(u)\) must be \(0\) for all points on the unit circle.

If \(g \neq 0\) for any unit vector, the limit depends on the path, and \(f\) fails to be differentiable at \((0,0)\). ∎

Evaluate \(f(0)\)
Substitute \(x = 0\) into the given condition: \[ \vert f(0) \vert \le \vert 0 \vert^2 \implies \vert f(0) \vert \le 0 \]
Since the absolute value must be non-negative (\(\vert f(0) \vert \ge 0\)), it follows that \(f(0) = 0\).

Definition of differentiability
A function \(f\) is differentiable at \(a = 0\) if there exists a linear transformation \(L\) such that: \[ \lim_{x \to 0} \frac{\vert f(x) - f(0) - L(x) \vert}{\vert x \vert} = 0 \]

Propose the zero transformation as the derivative candidate: \(L(x) = 0\).

Evaluate the limit expression
Substitute \(f(0) = 0\) and \(L(x) = 0\) into the limit definition: \[ \lim_{x \to 0} \frac{\vert f(x) - 0 - 0 \vert}{\vert x \vert} = \lim_{x \to 0} \frac{\vert f(x) \vert}{\vert x \vert} \]

Apply the Squeeze Theorem
Using the given condition for \(x \neq 0\): \[ 0 \le \vert f(x) \vert \le \vert x \vert^2 \]

Divide the entire inequality by \(\vert x \vert\): \[ 0 \le \frac{\vert f(x) \vert}{\vert x \vert} \le \frac{\vert x \vert^2}{\vert x \vert} \] \[ 0 \le \frac{\vert f(x) \vert}{\vert x \vert} \le \vert x \vert \]

Take the limit as \(x \to 0\) for the bounding terms: \[ \lim_{x \to 0} 0 = 0 \quad \text{and} \quad \lim_{x \to 0} \vert x \vert = 0 \]

Since both the lower and upper bounds approach \(0\), the Squeeze Theorem dictates: \[ \lim_{x \to 0} \frac{\vert f(x) \vert}{\vert x \vert} = 0 \]

Since the limit equals \(0\), the differentiability condition is perfectly satisfied by the linear map \(L(x) = 0\). Therefore, \(f\) is differentiable at \(x = 0\) and \(Df(0) = 0\). ∎