Question

Show that the sequence (f) where f(x)= x/(1+2nx^2), x∈ [1,∞[ is uniformaly convergent in [1,∞[

Answer 1

Answer on Question #69114 – Math – Real Analysis

Question

Show that the sequence $f_{n}(x) = \frac{x}{1 + 2nx^{2}}, x \in [1, \infty)$ is uniformly convergent in $[1, \infty)$.

Proof

Obviously the pointwise limit of $f_{n}(x)$ is 0. Indeed, for all fixed $x_0 \in [1, \infty)$

$$\lim _ {n \to \infty} f _ {n} (x _ {0}) = \lim _ {n \to \infty} \frac {x _ {0}}{1 + 2 n x _ {0} ^ {2}} = x _ {0} \lim _ {n \to \infty} \frac {1}{1 + 2 n x _ {0} ^ {2}} = \{1 + 2 n x _ {0} ^ {2} \to \infty \text{ as } n \to \infty \} = 0.$$

Now we apply the definition of the uniform convergence (see https://en.wikipedia.org/wiki/Uniform_convergence#Definition).

In our case the limit function is $f(x) \equiv 0$, and now we must find

$$a _ {n} = \sup _ {x \in [ 1, \infty)} | f _ {n} (x) - f (x) | = \sup _ {x \in [ 1, \infty)} \frac {x}{1 + 2 n x ^ {2}} > 0.$$

Applying a derivative we obtain:

$$\frac {d f _ {n} (x)}{d x} = \frac {1 + 2 n x ^ {2} - 4 n x ^ {2}}{(1 + 2 n x ^ {2}) ^ {2}} = \frac {1 - 2 n x ^ {2}}{(1 + 2 n x ^ {2}) ^ {2}} < 0, \text{ where } n \in \mathbb {N}, x \in [ 1, \infty).$$

Then $f_{n}(x) = \frac{x}{1 + 2nx^{2}}$ monotonically decreases in $x$ when $x \in [1, \infty)$, and

$$\sup _ {x \in [ 1, \infty)} f _ {n} (x) = \sup _ {x \in [ 1, \infty)} \frac {x}{1 + 2 n x ^ {2}} = f _ {n} (1) = \frac {1}{1 + 2 n}.$$

Since $a_{n} = \frac{1}{1 + 2n} \to 0$ as $n \to \infty$ (because $1 + 2n \to \infty$ as $n \to \infty$), we conclude that

$f_{n}(x) = \frac{x}{1 + 2nx^{2}}$ is uniformly convergent in $[1, \infty)$.

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