Question

Let \(x\) and \(y\) be elements of a normed linear space with \(\|x\|=1\) Show that either (a) \(\|x+\lambda y\| \geq 1\) for all \(\lambda>0\), or (b) \(\|x-\lambda y\| \geq 1\) for all \(\lambda>0\). Hint: Suppose that both are false and use the identity $$ \|\alpha a+(1-\alpha) b\| \leq|\alpha|\|a\|+\mid 1-\alpha\|b\| $$.

Short answer

Question: Show that for any two elements x and y of a normed linear space, with the condition that ∥x∥=1, either (a) ∥x+λy∥≥1 for all λ>0, or (b) ∥x−λy∥≥1 for all λ>0. Answer: This result is proven by contradiction by assuming that both (a) and (b) are false, then using the given hint and simplifying the inequality, we reach a contradiction, concluding that either (a) or (b) must be true.

Step-by-step solution

Define the base assumptions and contradiction

We will begin the proof by contradictory: Suppose that (a) and (b) are both false, meaning there exist \(\lambda_1,\lambda_2 > 0\) such that $$ \|x+\lambda_1 y\|<1 \quad \text{and} \quad \|x-\lambda_2 y\|<1. $$

Apply the hint's identity

Now let's use the hint by considering the following case: Set \(a = x - \lambda_2 y\) and \(b = x + \lambda_1 y\), and \(\alpha = \frac{\lambda_1}{\lambda_1 + \lambda_2}\). So we have \(\alpha > 0\) and \(1-\alpha>0\). Applying the given identity: $$ \|\alpha a+(1-\alpha) b\| \leq |\alpha|\|a\|+|1-\alpha|\|b\|. $$

Use the suppositions and complete the inequality

Now, substitute the expressions for \(a\), \(b\), and \(\alpha\). We obtain $$ \|\frac{\lambda_1}{\lambda_1+\lambda_2}(x-\lambda_2 y)+(1-\frac{\lambda_1} {\lambda_1+\lambda_2})(x+\lambda_1 y)\| \leq \frac{\lambda_1}{\lambda_1+\lambda_2}\|x-\lambda_2 y\|+\frac{\lambda_2}{\lambda_1+\lambda_2}\|x+\lambda_1 y\|. $$ Simplifying, we get $$ \|x\| \leq \frac{\lambda_1}{\lambda_1+\lambda_2}\|x-\lambda_2 y\|+\frac{\lambda_2}{\lambda_1+\lambda_2}\|x+\lambda_1 y\|. $$

Apply the base assumptions and obtain a contradiction

We know that \(\|x\| = 1\), and by our supposition, we have \(\|x-\lambda_2 y\|<1\) and \(\|x+\lambda_1 y\|<1\). Now, let's substitute these values into the inequality obtained in Step 3: $$ 1 \leq \frac{\lambda_1}{\lambda_1+\lambda_2}(1-\delta_2)+\frac{\lambda_2}{\lambda_1+\lambda_2}(1-\delta_1), $$ where \(\delta_1,\delta_2>0\) satisfy \(\|x-\lambda_2 y\|=1-\delta_2\) and \(\|x+\lambda_1 y\|=1-\delta_1\).

Simplify and reach the final contradiction

Simplify the inequality: $$ 1 \leq 1-\frac{\lambda_1}{\lambda_1+\lambda_2}\delta_2+\frac{\lambda_2}{\lambda_1+\lambda_2}\delta_1 $$ This inequality can be written as: $$ 0 \leq -\frac{\lambda_1\delta_2-\lambda_2\delta_1}{\lambda_1+\lambda_2}. $$ However, since the numerators and denominators are all positive, this inequality is in contradiction with our initial assumption, and we conclude that at least either (a) \(\|x+\lambda y\| \geq 1\) for all \(\lambda > 0\) or (b) \(\|x-\lambda y\| \geq 1\) for all \(\lambda > 0\) is true.

Key concepts

The Triangle Inequality

At the heart of many problems in normed linear spaces is the triangle inequality, a fundamental theorem that underpins the structure of such spaces.

Simply put, in any normed linear space, the triangle inequality states that for any vectors \(a\) and \(b\), the norm of their sum is less than or equal to the sum of their norms. Mathematically, this is represented as \[ \|a+b\| \leq \|a\| + \|b\| \].

Our problem relies on a variant of the triangle inequality, which can also be applied in the case of weighted sums of vectors. This is crucial when proving statements about linear combinations in normed spaces, as in the exercise provided. The triangle inequality ensures that the entire set of linear combinations of vectors (\