problem 2. let f ∈ c¹(ℝ), f(x + 1) = f(x), for all x, then we have ||f||∞ ≤ ∫₀¹|f(t)|dt + ∫₀¹|f(t)|dt…

problem 2. let f ∈ c¹(ℝ), f(x + 1) = f(x), for all x, then we have ||f||∞ ≤ ∫₀¹|f(t)|dt + ∫₀¹|f(t)|dt. solution. from the mean - value theorem there exists ξ ∈ 0,1 such that f(ξ) = ∫₀¹f(t)dt. now f(t) - f(ξ) = ∫_ξ^t f(s)ds, then we have that |f(t) - f(ξ)| ≤ ∫₀¹|f(s)|ds. and |f(t)| ≤ |f(ξ)| + ∫₀¹|f(s)|ds for all t ∈ 0,1 which gives the desired estimate.

problem 2. let f ∈ c¹(ℝ), f(x + 1) = f(x), for all x, then we have ||f||∞ ≤ ∫₀¹|f(t)|dt + ∫₀¹|f(t)|dt. solution. from the mean - value theorem there exists ξ ∈ 0,1 such that f(ξ) = ∫₀¹f(t)dt. now f(t) - f(ξ) = ∫_ξ^t f(s)ds, then we have that |f(t) - f(ξ)| ≤ ∫₀¹|f(s)|ds. and |f(t)| ≤ |f(ξ)| + ∫₀¹|f(s)|ds for all t ∈ 0,1 which gives the desired estimate.

Answer

Explanation:

Step1: Apply mean - value theorem

By the mean - value theorem for integrals, since (f\in C^1(\mathbb{R})) and (f(x + 1)=f(x)) for all (x), there exists (\xi\in[0,1]) such that (f(\xi)=\int_{0}^{1}f(t)dt).

Step2: Use fundamental theorem of calculus

We know that (f(t)-f(\xi)=\int_{\xi}^{t}f^{\prime}(s)ds) by the fundamental theorem of calculus. Then (|f(t)-f(\xi)|\leq\int_{0}^{1}|f^{\prime}(s)|ds).

Step3: Obtain the inequality for (|f(t)|)

From (|f(t)-f(\xi)|\leq\int_{0}^{1}|f^{\prime}(s)|ds), we can get (|f(t)|\leq|f(\xi)|+\int_{0}^{1}|f^{\prime}(s)|ds) for all (t\in[0,1]). Since (|f|{\infty}=\sup{t\in[0,1]}|f(t)|), and (|f(\xi)|=\left|\int_{0}^{1}f(t)dt\right|\leq\int_{0}^{1}|f(t)|dt), we have (|f|{\infty}\leq\int{0}^{1}|f(t)|dt+\int_{0}^{1}|f^{\prime}(t)|dt).

Answer:

The proof shows that (|f|{\infty}\leq\int{0}^{1}|f(t)|dt+\int_{0}^{1}|f^{\prime}(t)|dt) for (f\in C^1(\mathbb{R})) and (f(x + 1)=f(x)) for all (x).