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Operators on Infinite Dimensional Vector Spaces 2

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Simple Examples of Linear Bounded Operators

Example 1.13: Let X,YX,Y be Banach spaces and assume that XX is finite-dimensional. Let T:X→YT:X \rightarrow Y be a linear operator. Then TT is bounded.

Indeed, since all norms in XX are equivalent, it suffices to consider the case X=CnX = \mathbb{C}^n with the Euclidean norm ∥⋅∥2\|\cdot\|_2. Let e1,...,ene_1,...,e_n be the standard basis in Cn\mathbb{C}^n. For x∈Cnx \in \mathbb{C}^n, we have:

∥Tx∥Y=∥T∑i=1nxiei∥Y=∥∑i=1nxiTei∥Y≤∑i=1n∣xi∣∥Tei∥Y≤C∥x∥2,\|Tx\|_Y = \|T\sum_{i=1}^n x_i e_i \|_Y = \|\sum_{i=1}^n x_i Te_i\|_Y \leq \sum_{i=1}^n |x_i| \|Te_i\|_Y \leq C \|x\|_2,

where C2=∥Te1∥Y2+⋯+∥Ten∥Y2C^2 = \|Te_1\|^2_Y + \cdots + \|Te_n\|^2_Y.

Example 1.14: Let t1,t2,...t_1,t_2,... be a bounded sequence of complex numbers. For 1≤p≤∞1 \leq p \leq \infty, let T:ℓp→ℓpT : \ell^p \rightarrow \ell^p be defined by:

T:(x1,x2,...)↦(t1x1,t2x2,...).T: (x_1,x_2,...) \mapsto (t_1x_1,t_2x_2,...).

Then TT is bounded and ∥T∥=sup⁡n≥1∣tn∣.\|T\| = \sup_{n\geq 1} |t_n|.

Example 1.15: Let Ω\Omega be any measurable space, and let t∈L∞(Ω)t \in L^{\infty}(\Omega). Then:

T:f(x)↦t(x)f(x)T: f(x) \mapsto t(x)f(x)

defines a bounded operator on Lp(Ω)L^p(\Omega) (for any p≥1p \geq 1) and ∥T∥=∥t∥∞\|T\| = \|t\|_{\infty}.

Example 1.16: Let S:ℓ2→ℓ2S : \ell^2 \rightarrow \ell^2 be the shift operator:

S:(x0,x1,x2,...)↦(0,x0,x1,...).S: (x_0,x_1,x_2,...) \mapsto (0,x_0,x_1,...).

Then SS is bounded and ∥S∥=1\|S\| = 1. One can also consider the backwards shift operator S∗:ℓ2→ℓ2S^* : \ell^2 \rightarrow \ell^2 (the reason for the notation will become clear in the next section),

S∗:(x0,x1,x2,...)↦(x1,x2,...).S^* : (x_0,x_1,x_2,...) \mapsto (x_1,x_2,...).

Then we also have ∥S∗∥=1\|S^*\| = 1.

Example 1.17: Let a∈Ra \in \mathbb{R} and let Ta:Lp(R)→Lp(R)T_a : L^p(\mathbb{R}) \rightarrow L^p(\mathbb{R}) be defined by:

Ta:f(x)↦f(x−a).T_a : f(x) \mapsto f(x-a).

Then TaT_a is bounded and ∥Ta∥=1\|T_a\| = 1.