Consider the Banach space V = C ( [ − 1 , 1 ] ) V = C\p{\br{-1, 1}} V = C ( [ − 1 , 1 ] ) [ − 1 , 1 ] \br{-1, 1} [ − 1 , 1 ]
∥ f ∥ = sup { ∣ f ( x ) ∣ ∣ x ∈ [ − 1 , 1 ] } for f ∈ V . \norm{f} = \sup\,\set{\abs{f\p{x}} \mid x \in \br{-1, 1}} \quad\text{for } f \in V. ∥ f ∥ = sup { ∣ f ( x ) ∣ ∣ x ∈ [ − 1 , 1 ] } for f ∈ V . Let B = { f ∈ V ∣ ∥ f ∥ ≤ 1 } B = \set{f \in V \mid \norm{f} \leq 1} B = { f ∈ V ∣ ∥ f ∥ ≤ 1 } V V V
Show that there exists a bounded linear functional Λ : V → R \func{\Lambda}{V}{\R} Λ : V → R Λ ( B ) \Lambda\p{B} Λ ( B ) R \R R
Solution.
Define
Λ ( f ) = ∫ 0 1 f d x − ∫ − 1 0 f d x , \Lambda\p{f} = \int_0^1 f \,\diff{x} - \int_{-1}^0 f \,\diff{x}, Λ ( f ) = ∫ 0 1 f d x − ∫ − 1 0 f d x , i.e., Λ \Lambda Λ φ ( x ) = − 1 \phi\p{x} = -1 φ ( x ) = − 1 [ − 1 , 0 ) \pco{-1, 0} [ − 1 , 0 ) φ ( x ) = 1 \phi\p{x} = 1 φ ( x ) = 1 ( 0 , 1 ] \poc{0, 1} ( 0 , 1 ] f ∈ B f \in B f ∈ B
∣ Λ ( f ) ∣ ≤ ∫ 0 1 ∥ f ∥ d x + ∫ − 1 0 ∥ f ∥ d x = 2. \abs{\Lambda\p{f}}
\leq \int_0^1 \norm{f} \,\diff{x} + \int_{-1}^0 \norm{f} \,\diff{x}
= 2. ∣ Λ ( f ) ∣ ≤ ∫ 0 1 ∥ f ∥ d x + ∫ − 1 0 ∥ f ∥ d x = 2. Since B B B λ \lambda λ λ ( B ) \lambda\p{B} λ ( B ) ε > 0 \epsilon > 0 ε > 0
f ( x ) = { − 1 if − 1 ≤ x ≤ − ε x ε if − ε ≤ x ≤ ε 1 if ε ≤ x ≤ 1. f\p{x}
=
\begin{cases}
-1 & \text{if } -1 \leq x \leq -\epsilon \\
\frac{x}{\epsilon} & \text{if } -\epsilon \leq x \leq \epsilon \\
1 & \text{if } \epsilon \leq x \leq 1.
\end{cases} f ( x ) = ⎩ ⎨ ⎧ − 1 ε x 1 if − 1 ≤ x ≤ − ε if − ε ≤ x ≤ ε if ε ≤ x ≤ 1. Then f f f ∥ f ∥ = 1 \norm{f} = 1 ∥ f ∥ = 1
Λ ( f ) = ∫ − 1 − ε d x − ∫ − ε 0 x ε d x + ∫ 0 ε x ε d x + ∫ ε 1 d x = 1 − ε + ε 2 + ε 2 + 1 − ε = 2 − ε . \begin{aligned}
\Lambda\p{f}
&= \int_{-1}^{-\epsilon} \,\diff{x} - \int_{-\epsilon}^0 \frac{x}{\epsilon} \,\diff{x} + \int_0^\epsilon \frac{x}{\epsilon} \,\diff{x} + \int_\epsilon^1 \,\diff{x} \\
&= 1 - \epsilon + \frac{\epsilon}{2} + \frac{\epsilon}{2} + 1 - \epsilon \\
&= 2 - \epsilon.
\end{aligned} Λ ( f ) = ∫ − 1 − ε d x − ∫ − ε 0 ε x d x + ∫ 0 ε ε x d x + ∫ ε 1 d x = 1 − ε + 2 ε + 2 ε + 1 − ε = 2 − ε . Replacing f f f − f -f − f inf B Λ = − 2 \inf_B \Lambda = -2 inf B Λ = − 2 sup B Λ = 2 \sup_B \Lambda = 2 sup B Λ = 2 Λ \Lambda Λ B B B
Suppose Λ ( f ) = 2 \Lambda\p{f} = 2 Λ ( f ) = 2 f ( x ) = 1 f\p{x} = 1 f ( x ) = 1 ( 0 , 1 ] \poc{0, 1} ( 0 , 1 ] x 0 ∈ ( 0 , 1 ] x_0 \in \poc{0, 1} x 0 ∈ ( 0 , 1 ] f ( x 0 ) < 1 f\p{x_0} < 1 f ( x 0 ) < 1 f ( x ) ≤ 1 − ε < 1 f\p{x} \leq 1 - \epsilon < 1 f ( x ) ≤ 1 − ε < 1 I I I δ > 0 \delta > 0 δ > 0 ( 0 , 1 ] \poc{0, 1} ( 0 , 1 ]
Λ ( f ) = ∫ I f d x + ∫ [ 0 , 1 ] ∖ I f d x − ∫ − 1 0 f d x ≤ ∫ I f d x + ∫ [ 0 , 1 ] ∖ I f d x + ∫ − 1 0 d x ≤ ( 1 − ε ) δ + 1 − δ + 1 = 2 − ε δ < 2. \begin{aligned}
\Lambda\p{f}
&= \int_I f \,\diff{x} + \int_{\br{0,1} \setminus I} f \,\diff{x} - \int_{-1}^0 f \,\diff{x} \\
&\leq \int_I f \,\diff{x} + \int_{\br{0,1} \setminus I} f \,\diff{x} + \int_{-1}^0 \,\diff{x} \\
&\leq \p{1 - \epsilon}\delta + 1 - \delta + 1 \\
&= 2 - \epsilon\delta \\
&< 2.
\end{aligned} Λ ( f ) = ∫ I f d x + ∫ [ 0 , 1 ] ∖ I f d x − ∫ − 1 0 f d x ≤ ∫ I f d x + ∫ [ 0 , 1 ] ∖ I f d x + ∫ − 1 0 d x ≤ ( 1 − ε ) δ + 1 − δ + 1 = 2 − ε δ < 2. By symmetry, we need f ( x ) = − 1 f\p{x} = -1 f ( x ) = − 1 [ − 1 , 0 ) \pco{-1, 0} [ − 1 , 0 ) f ( 0 ) f\p{0} f ( 0 ) f f f f f f f f f Λ ( B ) = ( − 2 , 2 ) \Lambda\p{B} = \p{-2, 2} Λ ( B ) = ( − 2 , 2 )