Fourier series coefficients and convergence

[stripes]

Homework Statement

Third question of the day because this assignment is driving me crazy:

Suppose that \left\{ f_{k} \right\} ^{k=1}_{\infty} is a sequence of Riemann integrable functions on the interval [0, 1] such that

\int ^{0}_{1} |f_{k}(x) - f(x)|dx \rightarrow 0 as k \rightarrow \infty.

Show that \hat{f} _{k} (n) \rightarrow \hat{f} (n) uniformly in n as k \rightarrow \infty

Homework Equations

The Attempt at a Solution

I can't seem to do this rigorously. I can only approach it intuitively. Since the integral of the absolute value tends to zero, I want to say that f_{k}(x) \rightarrow f(x). But I'm not sure how to show that. If f_{k}(x) \rightarrow f(x) is indeed true, then is it not trivial that \hat{f} _{k} (n) \rightarrow \hat{f} (n)? Furthermore, how would I show the convergence is uniform? Do I just have to use the epsilon definition? I also want to say for all epsilon greater than zero, there exists f_k(x) such that |f_{k}(x) - f(x)| < \epsilon since f_k converges to f. But I need to first show that f_k converges to f, using the fact that the integral of the absolute value of the difference between the two converges to zero! I am piecing it together but I don't know how to write it down in the form of an answer. Thanks in advance.

Edit: f hat is the Fourier coefficient, I guess in complex form.

 

[MathematicalPhysicist]

Well it's really simple:

|\hat{f_k}-\hat{f}|=|\int (f_k(x)-f(x))exp(-i n x) dx |

Now use triangle inequality, and the fact that |exp(-inx)|=1.

 

[stripes]

<br /> <br /> \int ^{1}_{0} |f_{k}(x) - f(x)|dx \geq | \int ^{1}_{0} f_{k}(x) - f(x)dx | = | \int ^{1}_{0} ( f_{k}(x) - \int ^{1}_{0} f(x) ) dx | = | \int ^{1}_{0} ( f_{k}(x) (e^{-inx}) - \int ^{1}_{0} f(x) (e^{-inx}) ) dx |

because (e^{-inx}) = 1 for all n, x \geq 0. We know n starts at 1, and the question tells us x \in [0, 1].

<br /> <br /> | \int ^{1}_{0} ( f_{k}(x) (e^{-inx}) - \int ^{1}_{0} f(x) (e^{-inx}) ) dx | = | \hat{f} _{k}(n) - \hat{f} (n) |<br /> <br />

So by the comparison test, \int ^{0}_{1} |f_{k}(x) - f(x)|dx \rightarrow 0 as k \rightarrow \infty \Rightarrow | \hat{f} _{k}(n) - \hat{f} (n) | \rightarrow 0 \Rightarrow \hat{f} _{k}(n) \rightarrow \hat{f} (n)?

If this is right (I don't think my last statement is correct), do I just use the epsilon definition of uniform convergence by finding N so that |f_{k}(n) - f(n)| &lt; \epsilon?

 

[jbunniii]

<br /> <br /> \int ^{1}_{0} |f_{k}(x) - f(x)|dx \geq | \int ^{1}_{0} f_{k}(x) - f(x)dx | = | \int ^{1}_{0} ( f_{k}(x) - \int ^{1}_{0} f(x) ) dx | = | \int ^{1}_{0} ( f_{k}(x) (e^{-inx}) - \int ^{1}_{0} f(x) (e^{-inx}) ) dx |

because (e^{-inx}) = 1 for all n, x \geq 0.

No, it is not true that e^{-inx} = 1. It is true if you add absolute value signs: |e^{-inx}| = 1. Try starting like this:
$$\left|\hat{f}_k - \hat{f}\right| = \left|\int_0^1 f_k(x) e^{-inx} dx - \int_0^1 f(x) e^{-inx} dx\right| = \left|\int_0^1 (f_k(x) - f(x))e^{-inx} dx\right| \leq \int_0^1 \left| (f_k(x) - f(x))e^{-inx} \right| dx = \ldots$$

 

[stripes]

My mistake.

<br /> <br /> \ldots = \left|\int_0^1 (f_k(x) - f(x))e^{-inx} dx\right| \leq \int_0^1 \left| (f_k(x) - f(x))e^{-inx} \right| dx = \int ^{1}_{0} |f_{k} (x) - f(x)||e^{-inx}| dx = \int ^{1}_{0} |f_{k} (x) - f(x)|(1) dx which converges to zero. By the comparison test, \left|\hat{f}_k - \hat{f}\right| also converges to zero.

Then | \hat{f} _{k}(n) - \hat{f} (n) | \rightarrow 0 \Rightarrow \hat{f} _{k}(n) \rightarrow \hat{f} (n)?

 

[jbunniii]

Then | \hat{f} _{k}(n) - \hat{f} (n) | \rightarrow 0 \Rightarrow \hat{f} _{k}(n) \rightarrow \hat{f} (n)?

Yes, that's certainly true. If the absolute value of some quantity is approaching zero, then the quantity itself must also approach zero.

 

[stripes]

Alright. Thanks for your help thus far. I have shown that | \hat{f} _{k}(n) \rightarrow \hat{f} (n), but I need to show the convergence is uniform. My question is, will my conclusion help me show uniform convergence? I.e., can I work with what I've gotten to show uniform convergence, or must I start from scratch with the definition?

 

[jbunniii]

Alright. Thanks for your help thus far. I have shown that | \hat{f} _{k}(n) \rightarrow \hat{f} (n), but I need to show the convergence is uniform. My question is, will my conclusion help me show uniform convergence? I.e., can I work with what I've gotten to show uniform convergence, or must I start from scratch with the definition?

Your proof already shows that the convergence is uniform, because your upper bound, \int ^{1}_{0} |f_{k} (x) - f(x)| dx, does not depend on n.

 

[stripes]

Your proof already shows that the convergence is uniform, because your upper bound, \int ^{1}_{0} |f_{k} (x) - f(x)| dx, does not depend on n.

Right! Well it wasn't really my proof. You and MathematicalPhysicist basically did it. While I do understand it, I (obviously) have a hard time getting things started. I still have another question I haven't done and the other one you're helping me with!

Thanks again.