What's amiss in this Fourier convergence analysis?

Question

I worked out this solution to this basic PDE Fourier series convergence problem, but I suspect the result is "too easy to be correct," because all the answers point to no restriction on either $m$ and $b$:

Given this piecewise function: $$f(x) = \begin{cases} e^x, &-1 \leq x \leq 0,\\ mx + b, &0 \leq x \leq 1. \end{cases}$$ Without computing out any Fourier coefficients, find out the followings:
(a) for waht values of $m$ and $b$, if there is any, will the full Fourier series of $f(x)$ converge pointwise on $-1 \leq x \leq 1$?
(b) Same as the above, for converge uniformly?
(c) For converge in the $L^2$ sense?

Here are my answers on each of them:

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(a) My text says that if $f(x)$ is piecewise smooth on an interval, then its Fourier series converges pointwise on the interval. Piecewise smooth means both $f(x)$ and $f'(x)$ are piecewise continuous. Here, it is easy to see that $f(x)$ and $f'(x)$ are piecewise continuous for all real $m$'s and $b$'s. Thus there is no restriction on $m$ and $b$.

So far so good!

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(b) My text also says uniform convergence demands more stringent conditions: $f(x)$ has to have a continuous periodic extension, besides that $f'(x)$ has to be piecewise continuous on the interval. Here, I wrote that $f(x) = e^x$ has a continuous even function in $e^{-x}$ and $f(x) = mx + b$ also has continuous shift function, thus there is no restriction again for both $m, b$ in part (b).

Now I am beginning to doubt myself: Did I miss anything here?

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(c) Finally my text says that $f(x)$ converges in the $L^2$ sense if and only if $\int_{-l}^{l} | f(x) |^2 dx < \infty$. From here, we can see that both $\int_{-1}^{1} | e^x |^2 dx$ and $\int_{-1}^{1} | mx + b |^2 dx$ are finite. Thus again there is no restriction on both $m$ and $b$.

Now I am beginning to seriously second-guess myself: Is anything amiss here too?

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Thank you for your time and help.

Answer 1

You are right about $L^2$ convergence. For pointwise convergence, you need $b=1$ otherwise is not continuos in $0\ (\diamond)$. For uniform convergence, you need to make sure that the function $f$ can be made periodically continuos; that is, you have to connect the function at its extremes. Since $f(1) = m + b = m + 1$, and $f(-1) = e^{-1}$, you need $e^{-1} = m + 1 \implies m = e^{-1} - 1$

$\ \ \ \ \ (\diamond)$ In general, if your function is piecewise smooth, then the fourier series converges to $$\frac{f(x_0^+) + f(x_0^-)}2$$

$\ \ \ \ \ \ \ $So if you have a piecewise smooth function and you want pointwise convergence everywhere,$\ \ \ \ \ \ \ $ you need to make your function continuos

Answer 2

Observe that the function is not periodic. To make it periodic, I suggest change the value at the end point, i.e., define $f$ as

$f(x) = \left\{ \begin{array}{l} {e^x}{\rm{ , }} - 1 \le x \le 0,\\ mx + b,{\rm{ }}0 \le x < 1\\ {e^{ - 1}},{\rm{ }}x = 1. \end{array} \right .$

and extend by periodicity. Its period is of course 2.

Note that the function is plainly square integrable. Therefore, its Fourier series converges to $f$ in the L² norm.

Observe also that the function is of bounded variation. Therefore, its Fourier series converges everywhere. It converges to f(x) where f(x) is continuous, i.e., when x is not equal to $0$ or $-1$ or $1$. It converges to $(1+b)/2$ when $x = 0$; it converges to (e^-1+m+b)/2 when $x$ = 1 or -1.

If $b = 1 $ and e^-1 = $m+b$, then $f$ is a continuous function of bounded variation and so its Fourier series converges uniformly to $f$.

If $b$ is not equal to $1$ and e^-1 is not equal to $m+b$, then its Fourier series converges uniformly in any closed subinterval not containing {0, -1, 1} . To decide that it does not converge uniformly in the whole interval, one has to examine the local behaviour of $f$ at and near these three points.

If the Fourier series converges uniformly in $[-1, 1]$ to a function, then the function must be continuous in $[-1, 1]$. So in our case if $b$ is not equal to 1 or e^-1 is not equal to $m+b$, $f$ is not continuous at $0$ or at $-1$ or at $1$ and so the Fourier series cannot converge uniformly in $[-1, 1]$ .

See THeorem 23 to Theorem 29 in my article https://037598a680dc5e00a4d1feafd699642badaa7a8c.googledrive.com/host/0B4HffVs7117IbmZ2OTdKSVBZLVk/Fourier%20Series/Convergence%20of%20Fourier%20Series.pdf