# Fourier series coefficients and convergence

1. Jan 23, 2013

### stripes

1. The problem statement, all variables and given/known data

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 $k \rightarrow \infty$

2. Relevant equations

3. 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 fk(x) such that $|f_{k}(x) - f(x)| < \epsilon$ since fk converges to f. But I need to first show that fk 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.

2. Jan 23, 2013

### 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.

3. Jan 23, 2013

### stripes

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

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

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

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 $|f_{k}(n) - f(n)| < \epsilon$?

4. Jan 23, 2013

### jbunniii

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:
$$\left|\hat{f}_k - \hat{f}\right| = \left|\int_0^1 f_k(x) e^{-inx} dx - \int_0^1 f(x) e^{-inx} dx\right| = \left|\int_0^1 (f_k(x) - f(x))e^{-inx} dx\right| \leq \int_0^1 \left| (f_k(x) - f(x))e^{-inx} \right| dx = \ldots$$

5. Jan 23, 2013

### stripes

My mistake.

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

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

6. Jan 23, 2013

### jbunniii

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

7. Jan 23, 2013

### 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?

8. Jan 23, 2013

### jbunniii

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 $n$.

9. Jan 24, 2013

### stripes

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.