What can you say about a continuous function that is zero at all integer values?

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Let $f: \Bbb R \to \Bbb R $ be a continuous function such that $f(i) = 0 \ for \ all \ i \in \Bbb Z$. Then which of the following is true : –

A. Image(f) is closed in $\Bbb R$
B. Image(f) is open in $\Bbb R$
C. f is uniformly continuous
D. None of the above

For the correct option, look at the end of the question.

I tried to contradict the options by bringing examples(It was obviously the first step considering ‘None of the above was’ an option)

I wasn’t able to think of a single function that contradicts either of the options. However, graphically I would say that these functions satisfy the hypothesis and contradict the options
(Image uploaded)
enter image description here
Here’s what the functions are

1) Just like $sin x,$ the function touches 1 and -1. The major difference would be that this function touches the x-axis at all integral values and no where else.

2) The second graph touches the integers and zero and in each interval $(m,m+1)$, the graph’s maximum and minimum value tend to increase slightly and they tend to +1 and -1 as $x \to \infty$ Thus, the function gets arbitrarily close to 1 and -1 without touching them.

3) The third graph is just like $x Sin(x)$. The difference is that this one touches the x-axis at integer values. With each interval $(m,m+1)$, our graph has its ‘magnitude of difference’ increasing.

Are my graphs correct? If they, can they explicitly be expressed as familiar functions? If not, which explicitly express-able examples can be used to contradict the first three options?

The correct answer is D

Source – Tata Institute of Fundamental Research Graduate Studies Exam 2016

Solutions Collecting From Web of "What can you say about a continuous function that is zero at all integer values?"

Here’s one example that violates A,B,C:

$$f(x)= \frac{x^2}{1+x^2}(\sin^2(\pi x))^{1/(1+x^2)}.$$

The fraction out in front prevents $f$ from ever reaching $1,$ so $f(\mathbb R) = [0,1),$ which is neither open nor closed in $\mathbb R.$ The exponent insures that this function is not uniformly continuous. This is because

$$(\sin^2(\pi (n+h)))^{1/(1+(n+h)^2)} \to 1$$

so there will be sequences $x_n,y_n \to \infty$ such that $|x_n-y_n| \to 0$ such that $|f(x_n)-f(y_n)| \ge 1/2.$

By multiplying any polynomial $P(x)$ by $\sin^2(\pi x)$ (if you want to respect the sign of the polynomial, use $\sin(\pi x)$, if not), you can obtain those functions, which are (in radians):

Clic here for seeing the graph

The first is: $\sin(\pi x)$ (red)

The second one is almost : $\frac{x^2}{1+x^2}\sin(\pi x)$ (green) (I could not find a “nice” function here, thanks LutzL)

The third one is: $ax^2 \sin^2(\pi x)$ or $x \sin(\pi x)$ (blue)

Here is an example.