【归档】Prove that the set of continuous real-valued functions on the interval [0, 1] is a subspace...

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Prove that the set of continuous real-valued functions on the interval [0, 1] is a subspace of R[0,1]R^{[0, 1]}R[0,1]

Note Before Proof: 证明一个集合是另外一个集合的子空间，只要证明这个集合具有加法单位元0、加法封闭性、标量乘法封闭性即可。
Prove:
Let V = {f∣f:[0,1]→Rf | f: [0, 1] \rightarrow Rf∣f:[0,1]→R such that f is continuous}.
Part 1, additive identity（加法单位元）:
Take f0=0(∀x∈[0,1])f_{0} = 0 (\forall x \in [0, 1])f0=0(∀x∈[0,1]). Clearly f0f_{0}f0 is continuous and f0∈Vf_{0} \in Vf0∈V.
Part 2, closed under addition（加法封闭性）:
Take f,g∈Vf, g \in Vf,g∈V.
For each ϵ>0\epsilon > 0ϵ>0 and for each x∈[0,1]x \in [0, 1]x∈[0,1] there exists a δ>0\delta > 0δ>0. Such that if ∣x1−x2∣<δ|x_{1} - x_{2}| < \delta∣x1−x2∣<δ, then ∣f(x1)−f(x2)∣<ϵ2|f(x_{1}) - f(x_{2})| < \frac{\epsilon}{2}∣f(x1)−f(x2)∣<2ϵ, and ∣g(x1)−g(x2)∣<ϵ2|g(x_{1}) - g(x_{2})| < \frac{\epsilon}{2}∣g(x1)−g(x2)∣<2ϵ.
Since f+g=(f+g)(x)=f(x)+g(x)f + g = (f + g)(x) = f(x) + g(x)f+g=(f+g)(x)=f(x)+g(x), we have
∣(f+g)(x1)−(f+g)(x2)∣=∣[f(x1)−f(x2)]+[g(x1)−g(x2)]∣≤∣[f(x1)−f(x2)]∣+∣[g(x1)−g(x2)]∣<ϵ.\begin{aligned} & |(f + g)(x_{1}) - (f + g)(x_{2})|\\= & |[f(x_{1}) - f(x_{2})] + [g(x_{1}) - g(x_{2})]|\\ \le & |[f(x_{1}) - f(x_{2})]| + |[g(x_{1}) - g(x_{2})]|\\< & \epsilon.\end{aligned}=≤<∣(f+g)(x1)−(f+g)(x2)∣∣[f(x1)−f(x2)]+[g(x1)−g(x2)]∣∣[f(x1)−f(x2)]∣+∣[g(x1)−g(x2)]∣ϵ.
i.e. f+gf + gf+g is continuous at all x∈[0,1]x \in [0, 1]x∈[0,1].
Therefor, f+g∈Vf + g \in Vf+g∈V.
Part 3, closed under scalar multiplication（标量乘法封闭性）:
Take f∈Vf \in Vf∈V, and a∈Ra \in Ra∈R.
Assume a= 0, then
(af)(x)=a⋅f(x)=0,x∈[0,1](af)(x) = a \cdot f(x) = 0, x \in [0, 1](af)(x)=a⋅f(x)=0,x∈[0,1].
Clearly, afafaf is a continuous real-valued function on the interval [0, 1].
Assume a≠0a \neq 0a=0, for each ϵ>0\epsilon > 0ϵ>0 and for each x∈[0,1]x \in [0, 1]x∈[0,1], there exists a δ>0\delta > 0δ>0 such that if
∣x1−x2∣<δ|x_{1} - x_{2}| < \delta∣x1−x2∣<δ,
then
∣f(x1)−f(x2)∣<ϵa|f(x_{1}) - f(x_{2})| < \frac{\epsilon}{a}∣f(x1)−f(x2)∣<aϵ.
Now we have
∣(af)(x1)−(af)(x2)∣=∣a[f(x1)−f(x2)]∣=∣a∣⋅∣f(x1)−f(x2)∣<ϵ|(af)(x_{1}) - (af)(x_{2})| = |a[f(x_{1}) - f(x_{2})]| = |a|\cdot|f(x_{1}) - f(x_{2})| < \epsilon∣(af)(x1)−(af)(x2)∣=∣a[f(x1)−f(x2)]∣=∣a∣⋅∣f(x1)−f(x2)∣<ϵ.
Therefor, af∈Vaf \in Vaf∈V.
Thus V is a subspace of R[0,1]R^{[0, 1]}R[0,1].

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【归档】Prove that the set of continuous real-valued functions on the interval [0, 1] is a subspace...

Prove that the set of continuous real-valued functions on the interval [0, 1] is a subspace of R[0,1]R^{[0, 1]}R[0,1]

【归档】Prove that the set of continuous real-valued functions on the interval [0, 1] is a subspace...相关推荐

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