Question

Prove that the sequence {f_n(x)} where f_n(x)=nx/1+n^2x^2
is not uniformly convergent in [-1,1].

Answer 1

Answer to Question #85700 – Math – Real Analysis

Function is given as $f_{n}(x) = \frac{nx}{1 + n^{2}x^{2}}$

The value of function is zero when $x$ is zero for any value of $n$ including infinity

Which shows that the given function is constant value function?

Lets takes first derivatives of function

$$f _ {n} ^ {\prime} (x) = \frac {(n) (1 + n ^ {2} x ^ {2}) - n x (2 n ^ {2} x)}{(1 + n ^ {2} x ^ {2}) ^ {2}}$$ $$f _ {n} ^ {\prime} (x) = \frac {n - n ^ {3} x ^ {2}}{\left(1 + n ^ {2} x ^ {2}\right) ^ {2}}$$

In order to find out the critical point of the function we need to equate first derivative of function equal to zero

$$0 = \frac {n - n ^ {3} x ^ {2}}{\left(1 + n ^ {2} x ^ {2}\right) ^ {2}}$$

This gives us

$$x = \pm \frac {1}{n}$$

The value of function at this point is

$$f _ {n} \left(\pm \frac {1}{n}\right) = \pm \frac {1}{2}$$

But at the end point

$$x = \pm 1, \text { the functions have value of}$$ $$f _ {n} (\pm 1) = \frac {\pm 1}{1 + n ^ {2}}$$

In the set $[-1,1]$ , maxima and minima occur either at the end points or at the critical point.

Thus, in the interval $[-1,1]$ , the maximum value will be $\frac{1}{2}$ for all value of $n$ . So,

Limit $n$ tends to infinity for $\sup \left\{\langle f_n(x) - f(x)\rangle \langle x \in [-1,1]\right\} = \frac{1}{2} \neq 0$

So, $f_n$ does not converge uniformly to $f(x)$ on the $[-1,1]$.

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