Exercise-3

LADR

Exercise 1

Show that the set of differentiable real-valued functions \(f\) on the interval \((-4, 4)\) such that \(f^{\prime}(-1) = 3f(2)\) is a subspace of \(\mathbb{R}^{(-4, 4)}\).

Solution

Let us call this set \(S\).

Step-1

\(0 \in S\) as it satisfies the condition trivially.

Step-2

Let \(f, g \in S\). Then \(f^{\prime}(-1) = 3f(2)\) and \(g^{\prime}(-1) = 3g(2)\). Now: \[ \begin{aligned} (f + g)^{\prime}(-1) &= f^{\prime}(-1) + g^{\prime}(-1)\\ &= 3f(2) + 3g(2)\\ &= 3(f(2) + g(2))\\ &= 3(f + g)(2) \end{aligned} \] \(f + g \in S\). Therefore \(S\) is closed under addition.

Step-3

If \(f \in S\), then \(\lambda f \in S\). The argument is similar to step-2.

Footnotes

  1. Exercise-3, Exercises 1C, Page-24, Linear Algebra Done Right, Fourth Edition, Sheldon Axler↩︎