In algebra, a function is a relation between a set of inputs and a set of outputs under a condition that each input is related to exactly one output. There are many ways to describe a function.

Some functions may be defined by an algorithm or a formula that tells us how to find the output for particular input. And also by the graphs of function. Sometimes tables also reflect the output for a given input.

In mathematics we deal with different types of functions, like injective function, identity function, constant function, bounded function etc. A function having bounded set range is a bounded function. The range must have lower as well as upper bound.

Function can be bounded in real or complex field, if the set of its values is bounded. In this section we will study about bounded functions in detail with solved examples.

We know that a function has a domain and a range. If the range of a function is such that it is a finite closed interval on both sides, then such a function is called a bounded function.
A function is said to be bounded only if it is bounded in both sides, from above as well as from below. The two bounds are respectively called upper bound and lower bound.
For example see the graph of a function given below: 
Bounded Function
The domain is infinite in both the direction in the above function. However since the range is finite, we call this a bounded function.
We use the boundedness theorem to prove whether a given function is bounded or not. The theorem can be stated as follows:
A continuous function on a closed bounded interval is bounded.
Consider a function f that is defined and continuous at every point in the closed interval[a,b].
Assuming that f is not bounded above, we should have a point $x_1$ such that f($x_1$ )> 1, and another point $x_2$ such that $f(x_2 )$>2  and so on. If we now look at the sequence ($x_n$), then  using the Bolzano – Weierstrass theorem, we see that it has a subsequence ($x_ij$) that will converge to a point a  that lies in the interval [a,b]. We began by assuming that f is not bounded, so the sequence $(f(x_{ij}))$ is unbounded. However, since f is continuous, it has to converge to f(a). That means we have just contradicted ourselves. Therefore our basic assumption that f is unbounded is incorrect. That means f has to be bounded from above. 
Similar argument can be used to show that f is bounded from below as well. 
The upper bound of a bounded function is a constant that is just larger than the absolute value of any value of the function. Now suppose we have a family of functions.
This constant can be different for different functions in the same family. However, if we can find one single constant that bounds all the functions  in the given family of functions, then this family of functions is said to be uniformly bounded.
Example 1: Check weather sin x and cos x are bounded functions.

Solution: The trigonometric function sin(x) defined over the set of real numbers. Symbolically, f:R$\rightarrow$R,f(x)= Sin(x) is a bounded function. That is because we know for sure that the range of the sine function is [-1,1], thus making it a bounded function.
The graph of this function is as follows:

Sine Graph of Bounded Function

Cos Graph of Bounded Function
Just like sine, the cosine function is also a bounded function for the very same reasons. The function f:R$\rightarrow$ R,f(x)=cos(x) is also shown in the picture above. 

Now consider the function, f:R-{-1,1}$\rightarrow$ R, f(x) = $\frac{1}{x^2-1}$

Example 2: Show that rational function, f(x) = $\frac{1}{x^2-1}$ is bounded over all real values of x.

Solution: This function is defined for the given domain, that is all real numbers except -1 and 1.
Now let us sketch the graph of this function.

Equation Graph For Bounded Function

Note from the graph that the function has vertical asymptotes at x = 1 and x = -1, thus at these two points the graph of f(x) approaches negative and positive infinity.

Therefore this function cannot be said to be bounded. However, now if we restrict the domain as follows:

f:($-\infty$, -2)U (2,$\infty$)$\rightarrow$, f(x) = $\frac{1}{x^2-1}$

Then now the graph would look as follows:

Limits Graph For Bounded Function
This time we see that the function is now contained in the range [-$\frac{1}{3}$,$\frac{1}{3}$]. Thus now this function is bounded. 

The function f(x)= $\frac{1}{x^2-1}$ has no real vertical asymptotes. Thus this function is said to be bounded for all real values of x.