# Real Analysis Question, continuous functions over R with period 2π [closed]

Before getting into the question, just to let you guys know that I have a final tomorrow and this was on the past test but I'm lost as to how to proceed on this, so any help will be great!

Edit - I see people have said that this question needs additional content, i'm not sure what I can provide. As mentioned earlier, This is from a past final exam paper for a Real Analysis class, and was referred as good practice by the professor, though nothing else was said about it. I get that this question is a little broad in terms of the topic, but I'm looking for outright answers to the question but just the thinking behind the answer like Doug has mentioned below. I wouldn't be asking this if I had another option.

Consider a $$C^0_{2π-periodic}$$ of continuous functions over $$R$$ with period $$2π$$. Let us consider a map T which acts as a functions on $$C^0_{2π-periodic}$$ and produces new functions (T(f))(x) defined via the formula:

(T(f))(x) = $$\int_x^{x+π/6} f(t)dt$$

(a) Show that for every f $$\epsilon$$ $$C^0_{2π-periodic}$$, T(f) also lies in in the space $$C^0_{2π-periodic}$$.

(b)Thinking of the space $$C^0_{2π-periodic}$$ as a metric space (with the usual $$C^0$$ norm) show that the map T is continuous with respect to the $$C^0$$ metric on this space.

(c) Show that if f(x) is a trigonometric polynomial of order m, $$_f(x)$$ = $$\sum_{k=−m}^m$$ $$a_me^{im·x}$$ then T(f) is also a trigonometric polynomial of order m. Find an explicit expression for T(f) as a trigonometric polynomial in this case.

(d) Derive that there exists a constant C > 0 so that for every n $$\epsilon$$ Z, n ≠ 0 the $$n^{th}$$ Fourier coefficient $$\widehat{T(f)}$$ of T(f) satisfies |$$\widehat{T(f)}$$ (n)| ≤ _C|$$\hat{f}$$(n)| · $$|n|^{−1}$$ and |$$\widehat{T(f)}$$(0)| $$\le$$ C|$$\hat{f}$$(0)

(e) Consider the space B of 2$$\pi$$-periodic and Riemann-integrable functions R($$\Bbb{R}$$) for which $$\int_{-\pi}^{\pi}|f(x)|^2 dx \le 1$$

Consider the space of sequences

{$${....\widehat{T(f)}(-n-1),\widehat{T(f)}(n)},...,\widehat{T(f)}(0),...,\widehat{T(f)}(n),\widehat{T(f)}(n+1),...$$}

where $$f \varepsilon$$ B. Show the closure of this space in $$\ell^2 (\Bbb{Z})$$ is compact.

## closed as off-topic by Peter Foreman, Lord Shark the Unknown, YiFan, Mike Earnest, callculusApr 12 at 22:41

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• This looks like five questions. – Lord Shark the Unknown Apr 12 at 18:40
• @LordSharktheUnknown I wish that were the case, but this was given as one question with 5 parts – Yeti Apr 12 at 18:43
• But, which parts do you think you understand, and were do you still need help? How did you answer it on the previous test? Were there any notes on your answer? – Doug M Apr 12 at 18:55
• @DougM This wasn't on my previous test, just a previous year's test on the course. I'm looking for help on the entire question, preferably just a little help just on how to start and proceed with it. – Yeti Apr 12 at 20:09

To start with, if it is continuous and periodic, it can be modeled as a Fourier series. This is suggested in the later parts of the question.

a) is $$T(f(x))$$ periodic?

$$T(f(x+2\pi) = \int_{x+2\pi}^{x+\frac {13}{6}\pi} f(t)\ dt\\ u = t-2\pi\\ \int_{x}^{x+\frac {1}{6}\pi} f(u-2\pi)\ du\\ \int_{x}^{x+\frac {1}{6}\pi} f(u)\ du = T(f(x))$$

Is it continuous?

$$|T(f(x+\delta)) - T(f(x))| = |\int_{x}^{x+\delta} f(t)\ dt|+ |\int_{x+\frac 16 \pi}^{x+\frac 16 \pi + \delta} f(t)\ dt|$$

And can you show that these are less than delta.

Or you could look at the Fourier series.

$$\int_x^{x+\frac 16\pi} e^{imx}$$ continous and differentiable for all natural $$m$$

Sorry, I don't have time for a more complete answer, but does that get you started?

• That does help, and it does get me started. (b) shouldn't be a problem, for (c), we use the fact that T is continuous from (b) and it lies in the same space from (a), I'm not sure how to show it has the same order. For (d), well you just take the Fourier coefficients and play around the with the inequality till you get it? (e) I have no idea how to proceed. – Yeti Apr 13 at 0:10