Question

Let \(X\) be a normed linear space, and define the mapping \(u\) by \(u(x)=x /\|x\|\), for \(x \in X \backslash\\{\theta\\} .\) Show that $$ \|u(x)-u(y)\| \leq \frac{2\|x-y\|}{\|x\|} $$ Deduce that \(u\) is continuous on \(X \backslash\\{\theta\\}\).

Short answer

Question: Prove that the mapping \(u(x) = x / \|x\|\) is continuous on a normed linear space \(X \backslash \{\theta\}\), given that \(\|u(x) - u(y)\| \leq \frac{2\|x - y\|}{\|x\|}\). Solution: We showed that \(\|u(x) - u(y)\| \leq \frac{2\|x - y\|}{\|x\|}\) holds and used this result to deduce that \(u\) is continuous on \(X \backslash \{\theta\}\) by finding a \(\delta > 0\) such that \(\|x - y\| < \delta\) implies \(\|u(x) - u(y)\| < \varepsilon\) for any \(\varepsilon > 0\).

Step-by-step solution

Rewrite the inequality

Let's rewrite the given inequality as: $$ \|u(x) - u(y)\| = \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{2\|x - y\|}{\|x\|} $$

Use properties of the norm

Next, we will use properties of the norm to manipulate the expression: \begin{align*} \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| &= \left\|\frac{x\|y\|}{\|x\|\|y\|} - \frac{y\|x\|}{\|x\|\|y\|}\right\| \\ &= \frac{1}{\|x\|\|y\|}\|x\|y - y\|x\|\| \\ &= \frac{1}{\|x\|\|y\|}\|(x - y)\|x\|\|. \end{align*}

Apply the triangle inequality

Now, we apply the triangle inequality: $$ \|(x - y)\|x\|\| = \|x - y\|\|x\| \leq \|x - y\|(\|x\| + \|y\|). $$

Combine the results

Combining the results from Steps 2 and 3, we have $$ \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{1}{\|x\|\|y\|}\|x - y\|(\|x\| + \|y\|). $$ Removing the common term \(\|x\| + \|y\|\) in the inequality results in $$ \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| \leq \frac{2\|x - y\|}{\|x\|}. $$

Deduce the continuity of \(u\)

To show that \(u\) is continuous on \(X \backslash \{\theta\}\), let \(x, y \in X \backslash \{\theta\}\) and \(\varepsilon > 0\). We need to find a \(\delta > 0\) such that \(\|x - y\| < \delta\) implies \(\|u(x) - u(y)\| < \varepsilon\). From our inequality, we have $$ \|u(x) - u(y)\| \leq \frac{2\|x - y\|}{\|x\|}. $$ Let \(\delta = \varepsilon \|x\| \Rightarrow \|x - y\| < \delta\). Then $$ \|u(x) - u(y)\| \leq \frac{2\|x - y\|}{\|x\|} < \frac{2\delta}{\|x\|} = 2\varepsilon. $$ Thus, \(u\) is continuous on \(X \backslash \{\theta\}\).

Key concepts

Continuity in Normed Spaces

In mathematics, continuity is an essential concept that describes how functions behave as their inputs change incrementally. In the context of normed spaces, a function is continuous if small changes in input result in small changes in output. This is particularly important in functional analysis, where we frequently deal with spaces of infinite dimensions.

For a mapping like \( u(x) = \frac{x}{\|x\|} \), determining continuity involves checking whether for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( \|x-y\| < \delta \), then \( \|u(x) - u(y)\| < \varepsilon \). This ensures that no abrupt jumps occur in the function value as \( x \) changes slightly.

The exercise demonstrates continuity by showing that the expression is bounded by \( \frac{2\|x-y\|}{\|x\|} \), effectively linking changes in \( x \) and \( y \) through a bounded inequality. This bound decreases as \( \|x-y\| \) becomes smaller, formally proving \( u \) is continuous.

Triangle Inequality

The triangle inequality is a fundamental principle in mathematics, particularly useful in the study of vector spaces and norms. It indicates that for any vectors \(a\) and \(b\) in a normed space, the inequality \(\|a + b\| \leq \|a\| + \|b\|\) holds.

This property ensures that the sum of the lengths of any two sides of a triangle is always greater than or equal to the length of the third side. This concept is cleverly employed in various fields, such as geometry, computer science, and physics.

In our specific problem, the triangle inequality is used to simplify expressions involving vectors \((x - y)\) and their norms. By applying it, we are able to replace more complex terms with simpler, additive forms, which allows us to explicitly control the behavior of the function \( u \) as inputs change.

• Simplifies the manipulation of expressions involving sums of vectors.
• Facilitates understanding of distance and length relations in space.

Norm Properties

Norms are critical elements in the understanding of linear algebra and functional analysis. A norm is a function that assigns a non-negative length or size to each vector in a vector space, adhering to a few essential properties:

• Non-negativity: \( \|x\| \geq 0 \) and \( \|x\| = 0 \) if and only if \( x = 0 \)
• Scalar multiplicity: \( \|\alpha x\| = |\alpha| \|x\| \) for any scalar \( \alpha \)
• Triangle Inequality: \( \|x + y\| \leq \|x\| + \|y\| \)

These properties help define the structure of vector spaces and are instrumental in analyzing convergence, continuity, and other fundamental concepts.

In the exercise, we leverage these properties to break down and simplify the expression \( \|\frac{x}{\|x\|} - \frac{y}{\|y\|}\| \). Understanding and applying norm properties allows us to derive meaningful results, like the continuity of function \( u \), by managing and bounding vector magnitudes in consistent ways.