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In order to check wheter a transformation is linear and bounded i need to show that

$$T(ax +by) = aT(x) + bT(y)$$

But when the examples get harder, I have trouble doing so. For instance the following three questions stumpled me. I am only used to proving linearity to mappings on the form $x \mapsto f(x)$.

b) $$T: \ \mathbb{R} \to \mathbb{R}, \ \ t^3 \mapsto 3t^2$$

This is just derivation right? And derivation is a linear and bounded transformation..

d) $$T: \ P_n(\mathbb{R}) \to P_n(\mathbb{R}), \ \ \sum_{j=0}^n a_j x^j \mapsto \sum_{j=1}^n j a_j x^{j-1}$$

with $\| \sum a_j x^j\|_{P_{n}(\mathbb{R})} = \|(a_0, \ldots, a_n)\|_{\mathbb{R}^{n+1}}$.

f) $$C^1([0,1],\mathbb{R}) \cap BC([0,1],\mathbb{R})\to BC([0,1,\mathbb{R}]), \qquad f \mapsto f'$$ where the $BC([0,1],\mathbb{R})$-norm on the dense subspace $C^1([0,1],\mathbb{R})$ of $BC([0,1],\mathbb{R})$

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1 Answer 1

up vote 3 down vote accepted

(b) is not a derivation. It is not linear. For example, $1 \mapsto 3$, $-1 \mapsto 3$, but $(1 + -1) \mapsto 0 \ne 3 + 3$.

If $P_n(\mathbb R)$ means the ring of polynomial of degree not exceeding $n$ with coefficients in $\mathbb R$, then $P_n(\mathbb R)$ is isomorphic to $\mathbb R^{n+1}$ with an obvious isomorphism suggested by the metric. The transformation $T$ can be represented as a matrix, hence linear and bounded.

(f) The map is obviously linear. It is not bounded with respect to the uniform metric. For example, consider the family of functions

$$ f_a(x) = \sin(ax) $$

for $a \in \mathbb R$. This family is contained in a ball centered at the origin, but its image has no upper bound on the norm.

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