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Soit E l’ensemble des applications f : [0, +∞[→ R continues telles que l’int´egrale∫ +∞0f (t)2e−tdtconverge.1. Montrer que E est un R-espace vectoriel.

Question

Soit E l’ensemble des applications f : [0, +∞[→ R continue telles que l’intégale ∫0+∞f(t)2e−tdt \int_0^{+\infty} f(t)^2 e^{-t} dt converge.

  1. Montrer que E est un R-espace vectoriel.
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Solution

To show that the set E E of continuous functions f:[0,+∞[→R f : [0, +\infty[ \to \mathbb{R} for which the integral ∫0+∞f(t)2e−tdt \int_0^{+\infty} f(t)^2 e^{-t} dt converges is a vector space over R \mathbb{R} , we will follow these steps:

1. Break Down the Problem

We need to demonstrate that E E satisfies the properties of a vector space:

  1. It contains the zero vector (the zero function).
  2. It is closed under addition: If f1,f2∈E f_1, f_2 \in E , then f1+f2∈E f_1 + f_2 \in E .
  3. It is closed under scalar multiplication: If f∈E f \in E and c∈R c \in \mathbb{R} , then cf∈E c f \in E .

2. Relevant Concepts

  • Zero Function: The function f(t)=0 f(t) = 0 for all t t in [0,+∞[ [0, +\infty[ .
  • Convergence of Integrals: The integral ∫0+∞f(t)2e−tdt \int_0^{+\infty} f(t)^2 e^{-t} dt must be finite.

3. Analysis and Detail

  1. Zero Function:

    • Consider f(t)=0 f(t) = 0 . Then, ∫0+∞f(t)2e−tdt=∫0+∞02e−tdt=0 \int_0^{+\infty} f(t)^2 e^{-t} dt = \int_0^{+\infty} 0^2 e^{-t} dt = 0 This integral converges, thus 0∈E 0 \in E .

  2. Closure under Addition:

    • Let f1,f2∈E f_1, f_2 \in E . Then, ∫0+∞f1(t)2e−tdt<+∞and∫0+∞f2(t)2e−tdt<+∞ \int_0^{+\infty} f_1(t)^2 e^{-t} dt < +\infty \quad \text{and} \quad \int_0^{+\infty} f_2(t)^2 e^{-t} dt < +\infty We need to show that: ∫0+∞(f1(t)+f2(t))2e−tdt<+∞ \int_0^{+\infty} (f_1(t) + f_2(t))^2 e^{-t} dt < +\infty By the Cauchy-Schwarz inequality: (f1(t)+f2(t))2≤2(f1(t)2+f2(t)2) (f_1(t) + f_2(t))^2 \leq 2(f_1(t)^2 + f_2(t)^2) Therefore, ∫0+∞(f1(t)+f2(t))2e−tdt≤2(∫0+∞f1(t)2e−tdt+∫0+∞f2(t)2e−tdt)<+∞ \int_0^{+\infty} (f_1(t) + f_2(t))^2 e^{-t} dt \leq 2 \left( \int_0^{+\infty} f_1(t)^2 e^{-t} dt + \int_0^{+\infty} f_2(t)^2 e^{-t} dt \right) < +\infty Hence, f1+f2∈E f_1 + f_2 \in E .

  3. Closure under Scalar Multiplication:

    • Let f∈E f \in E and c∈R c \in \mathbb{R} . Then, ∫0+∞(cf(t))2e−tdt=c2∫0+∞f(t)2e−tdt \int_0^{+\infty} (c f(t))^2 e^{-t} dt = c^2 \int_0^{+\infty} f(t)^2 e^{-t} dt Since ∫0+∞f(t)2e−tdt<+∞ \int_0^{+\infty} f(t)^2 e^{-t} dt < +\infty , it follows that c2∫0+∞f(t)2e−tdt<+∞ c^2 \int_0^{+\infty} f(t)^2 e^{-t} dt < +\infty . Thus, cf∈E c f \in E .

4. Verify and Summarize

We have verified that:

  • The zero function is in E E .
  • E E is closed under addition.
  • E E is closed under scalar multiplication.

Final Answer

Therefore, E E is a vector space over R \mathbb{R} .

This problem has been solved

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