Let $f$ and $g$ be two positive valued functions defined on $[-1,1]$, such that $f(x)f(-x)=1$, and $g$ is an even function with

Question

Let $f$ and $g$ be two positive valued functions defined on $[-1,1]$, such that $f(x)f(-x)=1$, and $g$ is an even function with $\int _{-1}^{1}g(x)= 1$. Then $I=\int _{-1}^{1} f(x)g(x) dx$ satisfies

$$A.I\geq 1\quad \quad B.I\leq 1\quad \quad C. \frac13<I<3 \quad \quad D.I=1$$

My partial solution goes like this:

If we put $f(x)=e^x$ and $g(x)=|x|$ , then, both $f,g$ satisfies all the given conditions. Also, we find $$I=\int _{-1}^{1} f(x)g(x) dx=\int _{-1}^{0} f(x)g(x) dx+\int _{0}^{1} f(x)g(x) dx=\int _{-1}^{0} -e^xx dx+\int _{0}^{1} xe^xdx\geq 1.$$ This procedure eliminates, options $B$ and $D.$

Now, how to determine the correct option between $A.$ and $C.$ ? Also, in this casez I think, I was just "lucky enough" to find such an example for the functions $f,g$. But how to solve these problems in an mcq (Multiple Choice Question) test where one have a time constraint ? I dont get it.

Answer 1

As requested, I expand my comment into an answer. There is no requirement on the continuity of $f$, so we may consider:

$$ f(x) = \begin{cases} \frac{1}{N} & x < 0 \\ 1 & x = 0 \\ N & x > 0 \end{cases} $$

for some $N \in \mathbb{R}^+$. (Note: the value of $f$ at $x=0$ is determined by the conditions given.)

Then: $$\int_{-1}^{1} f(x)g(x) dx = \frac{1}{N} \int_{-1}^0 g(x) dx + N \int_0^1g(x)dx$$

$$=\frac{1}{2} \bigg(\frac{1}{N} + N \bigg)$$

For $N$ arbitrarily large (not even that large!), we observe that only $(A):I \ge1$ will hold.

I could not be considered an authority on this topic, but here is some advice. Generally, you want to extremise when finding counterexamples -- to make a certain quantity small or large (here, the value of $I$), or to find objects which possess or lack certain crucial qualities (e.g. continuity, boundedness).

To establish which feature(s) should be extremised, you have to collect some data, of course -- try a few different approaches, plug in some numbers, and see what comes out. In particular, if you stumble across something that almost works by chance, try to identify, if any exist, the salient feature(s) that it extremises. Once these are found, try the simplest extremiser possible. In this case, it was a locally constant function, which is trivial to integrate.

Don't be afraid to over-simplify either. Suppose the question had required that $f$ was continuous. Then, note that argument carries across with an effectively identical function which is made continuous by some appropriate interpolation on an interval $( - \varepsilon, \varepsilon)$. I hope you agree that it is not even necessary here to find a formula to see that this would work. In either case, the principle of extremisation is what is key.

Lastly, think about heuristic arguments. In analysis, if a quantity can be observed to be small, there is a decent chance it can be made arbitrarily small. (Intuition will help here; similarly if we replace "small" with "large", of course.) Thus, it can be helpful to introduce new variables e.g. $N$ as above. In practical terms, we are enabled to try a family of infinite functions all at once. With the previous work, we know not only that $I$ can be larger than $3$, but that it is unbounded.

Indeed, note that, in principle, we are not even concerned with the value of $N$ which makes $I$ exceed $3$.

Answer 2

Rather than setting $g(x)=|x|$ it is simpler to set $g(x)=\frac12$ if we are looking for counterexamples.

Taking $f(x) = e^x$ is one option, and gives $$\int_{-1}^1 f(x) g(x)\,dx = \int_{-1}^1 \frac{e^x}{2}\,dx = \frac{e - e^{-1}}{2} \approx 1.15.$$ But equally well, we can take $f(x) = e^{kx}$ for any $k$, and get $$\int_{-1}^1 f(x) g(x)\,dx = \int_{-1}^1 \frac{e^{kx}}{2}\,dx = \frac{e^{k} - e^{-k}}{2k}.$$ For large $k$, this can be made as large as we like, ruling out (B), (C), and (D).

Even without that example in mind, we can think about what the condition on $f(x)$ actually tells us. It does not prohibit $f$ from doing literally anything we like on $(0,1]$, as long as we define its behavior on $[-1,0)$ to satisfy $f(-x) = \frac1{f(x)}$. All we need is that $f(0)=1$. So we can make $\int_0^1 f(x)\,dx$ as large as we like, which leads to the integral in the question being large as well.