Properties of an infinitely differentiable function

Question

Let $f:\mathbb{R} \to (0,\infty)$ be an infinitely differentiable function with $\int_{-\infty}^\infty f(t)dt = 1$. Pick the correct statement(s) from below:

1. $f(t)$ is bounded.
2. $\lim_{|t| \to \infty} f'(t) = 0$.
3. There exists $t_0 \in \mathbb{R}$ such that $f(t_0) \ge f(t)$ for all $t \in \mathbb{R}$.
4. $f''(a)=0$ for some $a \in \mathbb{R}$.

I have an idea for option A. We can choose some bump functions with heights increasing but width decreasing so that the function is unbounded. But how to construct such examples

Answer 1

Before we construct suitable functions for evaluating each of the conditions, we introduce the function $g(x)$ as $$g(x)=\begin{cases} e^{-\cot^2\pi x}&,\quad x\in[0,1]\\ 0&,\quad x\notin [0,1] \end{cases}.$$ This function is non-negative and infinitely differentiable over $\Bbb R$ (to prove this fact, you only need to prove it for $x=0$, which is a simple task). Furthermore, the epigraph area of $g(x)$ is bounded, which we denote by $I_0$. Finally, it is worth finding $g'(x)$ as $$ g'(x)= \begin{cases} 2\pi \cot \pi x(1+\cot^2 \pi x)e^{-\cot^2\pi x}&,\quad x\in[0,1]\\ 0&,\quad x\notin [0,1] \end{cases}. $$

Now, we try to provide a specific representation of $f(t)$ in terms of $g(t)$ that evaluates conditions 1 through 3 as false. Then, we will show that condition 4 always holds true.

A viable candidate for $f(t)$ is as follows $$ f(t)=\frac{k}{1+t^2}+\sum_{n=0}^\infty a_ng(tb_n-nb_n), $$ where $k$ is a positive constant, $a_n$ is a positive sequence and $b_n$ is a sequence whose terms are all above $1$. It is easy to see that $f(t)$ is positive throughout $\Bbb R$, because of the term $\frac{k}{1+t^2}$. We can therefore write $$ \int_{-\infty}^\infty f(x)dx {= k\pi+\int_{-\infty}^\infty \sum_{n=0}^\infty a_ng(b_nx-nb_n)dx \\= k\pi+\sum_{n=0}^\infty a_n\int_{-\infty}^\infty g(b_nx-nb_n)dx \\= k\pi+\sum_{n=0}^\infty a_n\int_{-\infty}^\infty g(b_nx-nb_n)dx \\= k\pi+\sum_{n=0}^\infty I_0\frac{a_n}{b_n} } $$ and $$ f'(x)=-\frac{2kx}{(1+x^2)^2}+\sum_{n=0}^\infty a_nb_n g'(xb_n-nb_n), $$ which yields $\lim_{n\to \infty} f'(n+\frac{1}{4b_n})=\lim_{n\to \infty}4\pi e^{-1}a_nb_n$ and $f(n+\frac{1}{2b_n})=\frac{k}{1+(n+\frac{1}{2b_n})^2}+a_n$.

From this point, any arbitrary function $f(x)$ with $\lim_{n\to \infty} a_nb_n=\lim_{n\to \infty} a_n=\infty$ violates all of the conditions 1 through 3. Keep in mind that the constant $k$ and the sequences $a_n$ and $b_n$ should be chosen such that $\int_{\Bbb R} f(x)dx=1$. For example, consider $$ f(x)=\frac{1/(2\pi)}{1+x^2}+\sum_{n=0}^\infty I_0^{-1}2^{n-2}g(x\cdot 4^n-n\cdot 4^n), $$ where $I_0=\int_0^1 e^{-\cot^2\pi x}dx$. It is easy to check that $f(x)$ is unbounded and $\lim_{|t| \to \infty} f'(t) $ does not exist.

Condition 4 is the only one that holds true. To prove this, we assume by contrary that $f'(x)\ne 0$ for any $x\in\Bbb R$. Since $f''(x)$ is continuous, we should either have $f''(x)>0$ for all $x$ or $f''(x)<0$ for all $x$. These two cases are similar, so we only go with the case of $f''(x)>0$ for all $x$.

It is impossible to have $f'(x)=0$ for all $x$. Therefore, there exists some $a$ for which $$ { f(a)>0 \\f'(a)\ne 0 \\f''(a)>0. } $$ The case $f'(a)>0$ yields $$ \int_a^x f''(u)du=f'(x)-f'(a)>0\implies \int_a^x f'(u)du=f(x)-f(a)>xf'(a) $$ for $x>a$. Then $f(x)>f(a)+xf'(a)$ which is impossible since it gives $\int_{\Bbb R}f(x)dx=\infty$. The case $f'(a)<0$ can be rejected either.

Conclusion

Conditions 1 through 3 do not hold in general, whilst condition 4 does.

Answer 2

Paul Sinclair has already explained (1)-(3) in the comments, and Mostafa Ayaz has already written up a complete answer but perhaps I can add a few comments and simplifications: this link Are there other kinds of bump functions than $e^\frac1{x^2-1}$? provides many examples/explanations of smooth bump functions (meaning they are $0$ outside of a bounded/compact set).

Pick your favorite one, and call it $\varphi$, and suppose it is supported on $[-B,B]$ (for example, you could pick $\varphi(x) = e^{1/(x^2-1)}$ which is supported on $[-1,1]$). Its integral, say $I$, is some finite positive number. Now remember that I can rescale a function $[x \mapsto f(x)]$ by $[x \mapsto f(ax)]$ (which if you look at the graph has the behavior of contracting the $x$-axis by a factor of $a$), and shift a function $[x\mapsto f(x)]$ by $[x \mapsto f(x-b)]$ (which if you look at the graph has the behavior of moving the graph right by $b$ units). Putting these actions together, we have that $\varphi_1(x):=2\varphi(4(x-2B))$ is $\varphi$ but 4-times skinnier, 2-times taller (hence the integral is $\frac 12$ that of $\varphi$), and centered at $2B$ instead of $0$ (hence its support does not overlap with the support of $\varphi$).

More generally, repeating this process (4-times skinnier and 2-times taller than the previous --- hence area/integral half that of the previous --- and shifted to the right $2B$ units compared to the previous), we get for every positive integer $k$ the function $\varphi_k(x):=2^k \varphi(4^k(x-2Bk))$. And the function whose graphs looks like all these bumps all on one picture is exactly the sum $\Phi:=\varphi + \varphi_1 + \varphi_2 + \ldots$ (the infinite sum is well-defined because for each $x\in \mathbb R$, at most one $\varphi_i$ is non-zero). The integral of this sum function is $I + \frac 12 I+ \frac 14 I + \ldots = 2I$, which is still a finite positive number of course.

If we insist that the resulting function be strictly positive, we can always add a strictly positive function with finite integral, say a Gaussian $e^{-x^2}$. Then the function $\Phi(x)+e^{-x^2}$ is strictly positive and has finite integral $I'$, so normalizing we get the function $\frac{1}{I'} (\Phi(x)+e^{-x^2})$ that is a counterexample to (1)-(3).

---

Finally for (4), one can pursue the FTC route Mostafa Ayaz takes, or a more geometric approach: again, let us suppose for sake of contradiction that $f''$ is never $0$. Then because $f$ is smooth, $f''$ is continuous, so by the intermediate value theorem, if $f''$ takes different signs, then there is a point at which $f''$ is $0$; contradiction. So then $f$ is either convex or concave (this equivalence between infinitessimal information, i.e. second derivative being $\geq 0$ or $\leq 0$, vs. global information, i.e. the secant line definition of convex/concave, does take a little bit of work; see e.g. If $f$ is a twice differentiable convex function on $\mathbb R$ then show that $f''$ is non negative on $\mathbb R$.).

But concave/convex functions lie above/below their tangent lines (Convexity of a Function Implies Function Lies Above Any Tangent Line):

• if $f$ lies above its tangent lines, then if the tangent line $L(x)$ has non-zero slope, then the positive area under the line (i.e. the integral of $\max\{0,L(x)\}$ on $\mathbb R$) will be infinite; contradiction. So that means that every tangent line to $f$ must have zero-slope, i.e. $f'(x)=0$ for all $x\in \mathbb R$, which by Rolle's theorem implies that $f$ is constant. But constant functions either have integral on $\mathbb R$ equal to $0$ or $\pm \infty$; contradiction.
• if $f$ lies below its tangent lines, then if the tangent line $L(x)$ has non-zero slope, then the negative area under the line (i.e. the integral of $\min\{0,L(x)\}$ on $\mathbb R$) will be (negative) infinite; contradiction. Again this means $f'\equiv 0$, which we already said was a contradiction.

Finally, I remark that one can prove (4) even if $f$ is just twice differentiable; instead of using IVT in the case that $f''$ is continuous, one uses the fact that $f''$ is a derivative of a differentiable function, and hence satisfies the IVT even if it might not be continuous, by Darboux's theorem.