# $f(x)>0$ or $f(x)+f(x+1)>0$. Is it true $\int_{-\infty}^{\infty}f>0$

Suppose $$f$$ is a continuous function on $$\Bbb R$$ s.t $$f(x)>0$$ or $$f(x)+f(x+1)>0$$. Is it true $$\int_{-\infty}^{\infty}f>0$$?

My guess: It is true.

I was thinking in this way that we can partition $$\Bbb R$$ s.t $$f^+=\{x \in \Bbb R: f(x)>0\}$$ then $$f^-=\{x \in \Bbb R: f(x)<0\}$$ and $$f^0=\{x \in \Bbb R: f(x)=0\}$$.

Now observe for any point $$x \in f^-$$ we have $$f(x+1)>-f(x)>0$$. Now $$\int_{-\infty}^{\infty}f=\lim_{n \to \infty}\int_{-n}^nf=\lim_{n \to \infty}\lim_{k \to \infty}R(P_{kn},f)$$ where $$P_{kn}$$ is the partition of the interval $$[-n,n]$$ into intervals of length $$\frac 1k$$ and $$R(P_{kn},f)$$ is the Riemann Sum(observe one thing that $$f$$ being continuous is Riemann Integrable on $$[-n,n]$$. Moreover, we do not care whether $$\lim_{n \to \infty}\int_{-n}^nf$$ diverges or not. Even if it diverges we have to show that it takes $$+\infty$$.

Now if $$x\in f^- \cap [-n,n]$$ then it will be in one of the subinterval $$I_{kn} \subseteq [-n,n]$$(WLOG we consider that $$I_{kn} \subseteq f^-$$ and corresponding $$I_{kn}+1$$ would be positive and the sum of these two will be positive.

I am writing my intuition. Please tell me whether the argument is correct or not or if there is a much more precise way to prove or write this problem!!

• Do you mean that $(\forall x \in \mathbb{R}, \ f(x)>0) \lor (\forall x \in \mathbb{R} \ f(x)+f(x+1)>0)$, or do you mean, $\forall x \in \mathbb{R} (f(x)>0 \lor f(x)+f(x+1)>0)$ Commented Nov 29, 2019 at 3:23
• The second one. Commented Nov 29, 2019 at 4:21
• Are you assuming that $\int_{-\infty}^{\infty}f(x) \, dx$ exists? Otherwise there are counterexamples. Commented Nov 29, 2019 at 9:36

As far as I understand the question and from the clarification by MathematicsStudent1122, the statement is false.

Assumptions: $$f$$ is continuous and for every $$x\in {\mathbb R}$$, either $$f(x)>0$$ or $$f(x)+f(x+1)>0$$.

The following counter example establishes that even the improper Riemann integral may fail to exist.

Counter example: Let $$\epsilon>0$$. Define $$f(x)$$ by $$f(x)=\left\{\begin{array}{ccc}\frac{\epsilon}{1+x^2},&&x\leq 0\\ -2^k\epsilon,&&x=\frac{6k-5}4,k=1,2,\cdots\\ \epsilon,&&\frac{3k-2}2\leq x\leq \frac{3k-1}2,k=1,2,\cdots\\ (2^k+1)\epsilon,&&x=\frac {6k-1}4\end{array}\right.$$ and extend by joining line segments to make the function continuous. Pictorially the graph for $$x\geq 0$$ consists of wavelike pattern of shape $$V$$, a horizontal segment and a $$\Lambda$$ of "period" $$\frac 32$$. By construction the integral $$\int_{-\infty}^0f(x)~dx =\left.{\epsilon}\arctan(x)\right|_{-\infty}^0=\frac{\pi}2\epsilon,$$ which is under control (i.e. convergent). For each of the $$V$$ shapes, the integral is $$\int_{\frac{3k-3}2}^{\frac{3k-2}2}f(x)~dx =-\frac 12\cdot\frac 12(2^k+1)\epsilon+\frac 12\epsilon=\frac 12\epsilon(\frac 12-2^{k-1}),k=1,2,\cdots.$$ The integral of the horizontal segments is $$\int_{\frac{3k-2}2}^{\frac{3k-1}2}f(x)~dx =\frac 12\epsilon,$$ and the integral of each of the $$\Lambda$$ shapes is $$\int_{\frac{3k-1}2}^{\frac{3k}2}f(x)~dx =\frac 12\cdot\frac 12\cdot 2^k\epsilon+\frac 12\epsilon=\frac 12\epsilon(1+2^{k-1}),k=1,2,\cdots.$$ It follows that the integral of one wave (consisting of $$V$$, $$-$$, and $$\Lambda$$) is $$\frac 54\epsilon$$, so the integral of the first $$k$$ waves is $$\int_0^{\frac {3k}2}f(x)~dx =\frac 54k\epsilon\rightarrow \infty.$$ But the integral of the first $$k$$ waves plus a $$(k+1)$$st $$V$$ shape is $$\int_0^{\frac {3k+1}2}f(x)~dx =\frac 54k\epsilon+\frac 12\epsilon(\frac 12-2^k)\rightarrow -\infty.$$ This shows that the improper Riemann integral does not exist.

However if one assumes that the Lebesgue integral (allowing $$\pm \infty$$) exists, then the assertion $$\int_{-\infty}^{\infty}f(x)~dx >0$$ is true. This can be argued as follows. Let $$E=\{x|f(x)\leq 0\}$$ and $$E_1=\{x+1|x\in E\}.$$ Then by assumptions, one has $$f(x)>0,\forall x\in E_1$$$$\Rightarrow E_1\subset E^c.$$ Let now $$E_2=E^c\setminus E_1.$$ Then $$\int f(x)~dx =\int_E f(x)~dx +\int_{E^c}f(x)~dx$$ $$=\int_E f(x)~dx +\int_{E_1}f(x)~dx +\int_{E_2}f(x)~dx$$ $$=\int_E f(x)~dx+\int_E f(x+1)~dx +\int_{E_2}f(x)~dx$$ $$=\int_E(f(x)+f(x+1))~dx +\int_{E_2}f(x)~dx >0.$$ QED

Define $$B := \{x| f(x) \leq 0 \}$$, $$C := \{x| f(x-1) \leq 0 \}$$; note that $$C = B + 1$$ and $$B$$ and $$C$$ are disjoint, closed and of equal measure. Note

$$\int_{B \cup C} f = \int_B f + \int_{C} f$$. Now we note that for all $$d > 0$$ if $$B_d = B \cap[-d,d]$$ and $$C_{d} = C \cap [-d+1,d+1]$$

$$\int_{B_d}f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]} f(b_{d,k,n})$$

and

$$\int_{C_d}f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{C_d \cap [-d+1 + \frac{2dk}{n}, -d+1 + \frac{2d(k+1)}{n}]} f(c_{d,k,n})$$

where $$b_{d,k,n} = inf\{b|b \in B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}] \}$$

and $$c_{d,k,n} = inf\{c|c \in C_d \cap [-d+1 + \frac{2dk}{n}, -d+1 + \frac{2d(k+1)}{n}]\}$$.

Now observe that $$c_{d,k,n} = b_{d,k,n} +1$$ and

$$m(B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]) = m(C_d \cap [-d+1 + \frac{2dk}{n}, -d+1 + \frac{2d(k+1)}{n}])$$ thus we can write

$$\int_{C_d} f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]} f(b_{d,k,n} +1)$$

Hence

$$\int_{B_d}f + \int_{C_d}f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]} f(b_{d,k,n} +1)+f(b_{d,k,n}) > 0$$

Hence

$$\int_{B}f + \int_{C}f > 0$$

now $$f > 0$$ on $$(B \cup C)^c$$ hence

$$\int f = \int_{B\cup C}f + \int_{(B\cup C)^c}f > 0$$. Note here that the integral of $$f$$ is assumed to exist.