Is the function in the $L^2$-sobolev space $H^{\frac{1}{2}} \ [- \pi, \pi]$ of order 1?

My professor and as a consequence the rest of my class are saying that the specified function is not in $$L^2$$-sobolev space $$H^{\frac{1}{2}} \ [- \pi, \pi]$$ of order 1.

TLDR: Just skip ahead to $$(\star \star \star)$$ this initial part is an explanation of my professor's notation as well as proving some trivial conditions.

$$L^2$$-sobolev space $$H^{\frac{1}{2}} \ [- \pi, \pi]$$

In order to be in $$L^2$$-sobolev space $$H^{\frac{1}{2}} \ [- \pi, \pi]$$ of order 1 a function $$f(x)$$ must satisfy three conditions:

1. The function must be continuous on the interval $$[-\pi,\pi]$$
2. The function must satisfy $$f(-\pi) = f(\pi)$$
3. The function's derivative, $$f'(x)$$, must exist and be continuous on the interval $$[-\pi,\pi]$$ except at a finite number of points.
4. The function must satisfy: $$\int_{-\pi}^{\pi}(f'(x))^2dx$$ < $$\infty$$, ie the integral of the derivative of the function squared over the interval must be finite.

If all these conditions are satisfied then the function is in $$L^2$$-sobolev space $$H^{\frac{1}{2}} \ [- \pi, \pi]$$ of order 1, and therefore the function's Fourier series converges uniformly and absolutely on the interval.

Now the function I am dealing with is the following:

$$f(x) = \begin{cases} \frac{\sin x}{x} \ \ \text{for} \ \ x \neq 0 \\ \ \ \ 1 \ \ \ \ \text{for} \ \ x =0 \end{cases}$$

Now (1) is satisfied since the function is continuous everywhere except perhaps at $$x = 0$$ however taking the limit at 0 and then applying L'Hopital shows that the function approaches $$1$$ as $$x \rightarrow 0$$. By definition of continuity and the piece-wise condition that $$f(x) = 1 \ \ \text{for} \ \ x =0$$, that is:

$$\lim_{x \rightarrow 0} f(x) = f(0)$$

Therefore the function is continuous everywhere on the interval.

For (2) the condition is easily shown it is trivial.

For(3) the functions derivative exists it is:

$$f'(x) = \frac{x \cos x - \sin x}{x^2}$$

and it is continuous everywhere including $$x = 0$$ where the function is defined as:

$$f'(x) = \begin{cases} \frac{x \cos x - \sin x}{x^2} \ \ \text{for} \ \ x \neq 0 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \text{for} \ \ x =0 \end{cases}$$

Using L'Hopital it is shown just like it was done before that the derivative is continuous at $$x = 0$$ and therefore the derivative exists and is continuous on the interval.

($$\star \star \star$$)

Finally (4) (this is where the disagreement comes in) My professor says that the integral condition diverges to infinity his argument being that the integral can be written as: $$\int_{-\pi }^{\pi } \left(\frac{x \cos x-\sin x}{x^2}\right)^2 dx = \int_{-\pi }^{\pi } \frac{x^2 \cos^2(x)}{x^4}dx + \int_{-\pi }^{\pi } \frac{\sin^2(x)}{x^4}dx - 2 \int_{-\pi }^{\pi } \frac{ x \cos(x) \sin x}{x^4}dx$$

His argument is that if you take the limit of each integral as $$x$$ goes to $$0$$ each integral goes to infinity therefore the integral diverges.

I disagree, my argument being that let there be a function $$g(x)$$, where $$g(x) = (f'(x))^2$$ $$\rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ g(x) = \begin{cases} \left(\frac{x \cos x - \sin x}{x^2}\right)^2 \ \ \text{for} \ x \neq0 \\ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for} \ \ x = 0 \end{cases}$$

This function is clearly what is inside the integral, and it is continuous on the interval except perhaps at $$0$$, using 3 applications of L'Hopital we show that the function $$g(x)$$ is continuous everywhere on the interval including at $$x = 0$$ (this being done in similar fashion as before). Then from a previous question I posted and confirmed on SE I determined via the extreme value theorem that this function has a maximum on the interval, (doesn't go to $$\infty$$), is continuous (due to piece wise condition), and the integral is evaluated over finite bounds. Therefore the definite integral must be finite, and thus the 4th condition is satisfied and therefore the function is in $$L^2$$-sobolev space $$H^{\frac{1}{2}} \ [- \pi, \pi]$$ of order 1. Who is right? I believe my arguments are carefully constructed and follow logically, and I disagree with my professor's method since $$\infty + \infty - \infty$$ does not necessarily imply the function goes to infinity it says nothing of the 'finiteness' or lack thereof of the integral and also this argument does not take into account the piecewise nature of the function. Please an expert on this help me out and is the integral finite in this case, Thank you!

Also just for some graphical grounding in my argument (although not proof) you can also look at the graph of $$g(x)$$ and see that considering that if the function is continuous at $$x = 0$$ which it was shown (due to piece wise) the integral of $$g(x)$$ must be finite.

$g(x)$">

1 Answer

The problem with the integral is that the formula $$\int_a^b\,(f+g)\,dx=\int_a^b\,f\,dx+\int_a^b\,g\,dx$$ is only valid under the assumption that either $$\int_a^b\,f\,dx$$ or $$\int_a^b\,g\,dx$$ is finite. The finiteness assumption can only be dropped if both $$f$$ and $$g$$ are non-negative or nonpositive. This is not the case in the way that your professor is dividing the integral.

An equivalente way of viewing it, it is the same as arguing that $$\int_0^1\,x\,dx$$ is not finite because it equals $$\int_0^1\,(x-1/x)\,dx+\int_0^1\,1/x\,dx$$ where each summand diverges to infinity.

• Thanks for the great insight, I will definitely bring this up. However is my argument showing that the integral is finite correct, as done above? – QuantumPanda Nov 16 '18 at 1:06
• Your argument is correct, continuous functions on a compact interval have always a finite integral. – Josué Tonelli-Cueto Nov 20 '18 at 14:23