Exercise 6.1.3

Answers

As per definition, compute

⟨ f, g ⟩ = \int_{0}^{1} f (t) g (t) dt = \int td e^{t} = t e^{t} |_{0}^{1} - \int e^{t} dt = e - (e - 1) = 1,

∥ f ∥ = \int_{0}^{1} f^{2} = \frac{1}{3},

∥ g ∥ = \int_{0}^{1} g^{2} = \frac{1}{2} (e^{2} - 1),

and

∥ f + g ∥ = ∥ t + e^{t} ∥ = {(∥ f ∥^{2} + 2 ⟨ f, g ⟩ + ∥ g ∥^{2})}^{\frac{1}{2}}

= \frac{1}{9} + 2 + \frac{1}{4} (e^{4} - 2 e^{2} + 1) = \frac{9 e^{4} - 18 e^{2} + 85}{36} .

And the two inequalities hold since

1 \leq \frac{1}{3} \cdot \frac{1}{2} (e^{2} - 1)

and

\frac{9 e^{4} - 18 e^{2} + 85}{36} \geq \frac{1}{3} + \frac{1}{2} (e^{2} - 1) .

jephianlin

2011-06-27 00:00

||f|| = 1/sqrt(3) and ||g|| = sqrt(1/2(e^2-1)) since ||f||^2 = <f,f>

stdionysius • 2025-12-06