# Showing a function is differentiable on $\mathbb{R}^+$

Question: Let $f \ge 0$ be an integrable function on $\mathbb{R}$. Define $g(t) := \displaystyle \int_\mathbb{R} \cos(tx)\ f(x) \; dx$ for $t \ge 0$. Show $g$ is twice-differentiable $\iff \displaystyle \int_\mathbb{R} x^2 f(x) \; dx <\infty$.

Solution: $(\Rightarrow)$ Suppose $g$ is twice-differentiable so that $| g''(0)|<\infty$. Then, \begin{align*} |g''(0)| &= \left\vert \lim \limits_{h \to 0} \dfrac{g(h) -2g(0) +g(-h)}{h^2} \right\vert \\ &= \lim \limits_{h \to 0}\left\vert \dfrac{g(h) -2g(0) +g(-h)}{h^2} \right\vert \\ &= \lim \limits_{h \to 0} \int_\mathbb{R} \left\vert \dfrac{ \cos(hx) -2 + \cos(-hx)}{h^2} f(x) \right\vert \; dx \quad (^* \text{Wrong, see edit}^*)\\ &= \lim \limits_{h \to 0} \int_\mathbb{R} \left\vert \dfrac{2( \cos(hx) -1) }{h^2} \right\vert f(x) \; dx \\ &\ge \int_\mathbb{R} \liminf \limits_{h \to 0} \left\vert \dfrac{2( \cos(hx) -1) }{h^2} \right\vert f(x) \; dx \quad \text{(Fatou)} \\ &= \int_\mathbb{R} x^2f(x) \; dx. \end{align*}

$(\Leftarrow)$ This is where I am sort of stuck. I am doing a standard technique of showing a partial derivative is bounded so I can push the derivative inside the integral using LDCT (Will moving differentiation from inside, to outside an integral, change the result?)

First, note that $g(t)$ is integrable since $$\left\vert \int_\mathbb{R} \cos(tx) f(x) \; dx \right\vert \le \|f\|_{L^1(\mathbb{R})}<\infty$$

Also, $\left\vert \frac{\partial^2 }{\partial t^2} \cos(tx) f(x) \right\vert = |x^2 \cos(tx) f(x)| \le |x^2 f(x)|,$ which is integrable by assumption. So $g''$ exists, assuming $g'$ does.

But I can't show $g'$ exists because $\left\vert \frac{\partial }{\partial t} \cos(tx) f(x) \right\vert = |x\sin(tx) f(x)|$, which is $\le$ both $|xf(x)|$ and $|x^2 t f(x)|$, neither of which seem to help . . .

If I attack with $g'(t) = \lim \limits_{h \to 0} \dfrac{g(t+h)-g(t)}{h} = \lim \limits_{h \to 0} \displaystyle \int_\mathbb{R} \dfrac{\cos(x(t+h)) - \cos(xt)}{h}f(x) \; dx$, I can't make progress either.

Thanks for the help.

EDIT: Let me fix the $\Rightarrow$ direction. Basically all I need to do is note that $2 - 2 \cos(hx) \ge 0$, so I can still apply Fatou. \begin{align*} -g''(0) &= \lim \limits_{h \to 0} -\dfrac{g(h) -2g(0) +g(-h)}{h^2} \\ &= \lim \limits_{h \to 0} \int_\mathbb{R} \dfrac{ -\cos(hx) +2 - \cos(-hx)}{h^2} f(x) \; dx \\ &= \lim \limits_{h \to 0} \int_\mathbb{R} \dfrac{2( 1-\cos(hx)) }{h^2} f(x) \; dx \\ &\ge \int_\mathbb{R} \liminf \limits_{h \to 0} \dfrac{2(1-\cos(hx)) }{h^2} f(x) \; dx \quad \text{(Fatou)} \\ &= \int_\mathbb{R} x^2f(x) \; dx. \end{align*}

• Your Edit is great. That's where we see use hypothesis $f\ge 0.$ (In the other direction it's not needed.) – zhw. Aug 10 '16 at 1:50
• Excellent. Thanks for the help! – user288742 Aug 10 '16 at 2:02

In your proof of $\implies,$ you have a mistake in going from line 2 to line 3: You've moved the absolute values inside the integral, so $=$ should be $\le 0.$

In the other direction, we can use the mean value theorem:

$$\tag 1 |\cos ((t+h)x) - \cos (tx)| = |(-\sin c)\cdot (hx)| \le |hx|$$

Now

$$\frac{g(t+h)-g(t)}{h} = \int \frac{\cos ((t+h)x) - \cos (tx)}{h} f(x)\, dx.$$

In absolute value, the integrand on the right is $\le |xf(x)|$ by $(1).$ Because $x^2f(x) \in L^1,$ $xf(x) \in L^1.$ So the dominated convergence theorem gives

$$\tag 2 g'(t) = \int (- \sin (tx))x f(x)\,dx.$$

This argument can be repeated in differentiating $(2),$ this time using $x^2f(x)\in L^1$ directly. We get $g''(t) = \int (- \cos (tx))x^2 f(x)\,dx.$

• Nice, +1. I like this better than my answer :-) – Bungo Aug 9 '16 at 19:33
• Shoot I thought I had the $\Rightarrow$ direction figured out. Do you see any way I can salvage it? – user288742 Aug 9 '16 at 20:39
• Ah never mind it's a simple fix - will update original post. I did have one question though on your answer, you said because $x^2 f(x) \in L^1,$ $xf(x) \in L^1$. The fact that $f \in L^1$ is needed here, correct? Just wanted to make sure I'm not missing anything. – user288742 Aug 9 '16 at 20:43
• Correct on that question. – zhw. Aug 9 '16 at 21:33

If $|h| \leq 1$, then \begin{aligned} \left|\frac{\cos(x(t+h)) - \cos(xt)}{h}\right| &= \left|\frac{\cos(xt)(\cos(xh)-1) - \sin(xt)\sin(xh)}{h}\right| \\ &\leq \left|\cos(xt)\right| \left|\frac{\cos(xh) - 1}{h} \right| + \left| \sin(xt) \right| \left| \frac{\sin(xh)}{h}\right| \\ &\leq |\cos(xt)|\left|\frac{x^2 h}{2}\right| + \left|\sin(xt)\right||x| \\ &\leq |\cos(xt)|\left| \frac{x^2}{2} \right| + \left|\sin(xt)\right||x| \\ &\leq \frac{x^2}{2} + |x| \\ \end{aligned} so $$\left(\frac{x^2}{2} + |x|\right)|f(x)|$$ serves as a dominating function once we recognize that $|x||f(x)|$ is bounded by $|f(x)|$ for $|x| \leq 1$ and by $x^2|f(x)|$ for $|x| > 1$.

• You've give away a lot there. How will you obtain the existence of $g''(t)$ from that estimate? – zhw. Aug 9 '16 at 19:20
• @zhw. That may be so, I was only addressing the part which the OP was stuck on: how to show existence of $g'$. I didn't check that the rest would follow. – Bungo Aug 9 '16 at 19:28

We have $f \ge 0$, so:

$$\int_{\Bbb R} |x f(x)| dx = -\int_{-\infty}^{-1} xf(x)dx + \int_{-1}^1 xf(x) dx + \int_1^{\infty} xf(x) dx \le \int_{-\infty}^{-1} x^2 f(x) dx + \int_{-1}^1 |x f(x)|dx + \int_{1}^{\infty} x^2 f(x)dx \le \|f\|_{L^1} + \|x^2 f(x)\|_{L^1} < \infty$$

So, $x \mapsto xf(x) \in L^1$ and we are done.