Question

Show that if \(X\) is a normed linear space and \(x, y \in X\) with \(\|x\| \leq 1,\|y\| \leq 1\), then $$ \|\alpha x+(1-\alpha) y\| \leq 1, \quad 0 \leq \alpha \leq 1 $$.

Short answer

Question: Prove that the unit ball in a normed linear space is convex. Answer: To prove that the unit ball in a normed linear space is convex, we need to show that for any two points x and y in the space with their norms less than or equal to 1, and any value α between 0 and 1, the norm of the linear combination αx + (1 - α)y is less than or equal to 1. By applying the properties of the norm, linearity and given conditions, we can deduce that the unit ball is indeed convex.

Step-by-step solution

Use the properties of the norm

Given \(x,y\in X\) with \(\|x\|\leq 1\) and \(\|y\|\leq 1\), let's consider the norm of the linear combination \(\alpha x+(1-\alpha) y\). By definition of the norm, we know that it satisfies the triangle inequality. In this case, we apply the triangle inequality to the sum \(\alpha x+(1-\alpha) y\): $$ \|\alpha x + (1 -\alpha) y\| \leq \|\alpha x\| + \|(1-\alpha) y\|. $$

Apply linearity and scaling

Next, we need to take into account that the norm is a linear function. This means that, for any scalar \(\alpha\), we have: $$ \|\alpha x\| = |\alpha| \cdot \|x\| $$ For our problem, we have \(0\leq\alpha\leq 1\). So, we can express the norms as follows: $$ \|\alpha x\| = \alpha\|x\| \quad \text{and} \quad \|(1-\alpha)y\|=(1-\alpha)\|y\|. $$ Now, we can substitute these expressions back into the inequality that we obtained in Step 1: $$ \|\alpha x + (1 -\alpha) y\| \leq \alpha\|x\| + (1-\alpha) \|y\|. $$

Use the given conditions on the norms of x and y

We are given that \(\|x\|\leq 1\) and \(\|y\|\leq1\). Using this information, we can write: $$ \|\alpha x + (1 -\alpha) y\| \leq \alpha(1) + (1-\alpha)(1) $$ This simplifies to: $$ \|\alpha x + (1 -\alpha) y\| \leq \alpha + (1-\alpha) = 1 $$ which concludes the proof.

Key concepts

Triangle Inequality

Understanding the triangle inequality is fundamental when dealing with normed linear spaces. The triangle inequality states that for any vectors x and y in the space, the norm of their sum is less than or equal to the sum of their norms, formally expressed as:
\[ \|x + y\| \leq \|x\| + \|y\| \.\]
The name stems from the geometrical interpretation of this property in Euclidean space, which can be visualized as the fact that the length of any one side of a triangle is always less than the sum of the lengths of the other two sides.

In the context of the given exercise, where x and y are vectors with norms less than or equal to 1, we demonstrated how one can use the triangle inequality to show that for any weighted linear combination within certain constraints, the resulting norm is also bounded. Essentially, the triangle inequality ensures that the space behaves in a way that is intuitive from a geometrical standpoint, and it is a key tool for establishing many important results in analysis and linear algebra.

Linear Combination

A linear combination in linear algebra involves constructing new vectors by multiplying existing vectors (the operands) by scalars (the coefficients) and adding the results. The general formula for a linear combination of two vectors is:
\[ \alpha x + \beta y \.\]
The importance of linear combinations lies in their ability to create every possible vector in a vector space, assuming x and y are part of a basis for that space.

In our exercise, we were examining a specific type of linear combination of two vectors x, y that lie within the unit ball, meaning their norms do not exceed 1. The coefficients here are \alpha and 1-\alpha, which depend on the scalar \alpha, and fall within the closed interval from 0 to 1. The correct choice and interpretation of these coefficients were crucial for proving the norm of such combinations does not exceed 1, reflecting the properties of a convex set.

Properties of the Norm

The norm in a normed linear space provides a way to measure the 'size' or length of vectors. It's a function that assigns a non-negative length to each vector, and it must satisfy certain properties:

• Non-negativity: \(\|x\|\geq0\), with equality if and only if \(x=0\).
• Scalar Multiplication: \(\|\alpha x\|=|\alpha|\cdot\|x\|\), reflecting how the norm changes with scaling.
• Triangle Inequality: Already discussed, it's the property that the norm of a sum is less than or equal to the sum of the norms.

When solving the exercise, these properties enabled us to manipulate the expressions and establish the desired inequality. By acknowledging that the norm remains unchanged or reduces when multiplied by a scalar between 0 and 1, we could assert that the norm of any convex combination of the elements within the set does not exceed 1. This insight is particularly useful in various applications including optimization and analysis.