Does sequence converge in $C([0,1],\mathbb{R})$?

Question

Does the sequence $(1-x)x^k$ converge uniformly on $[0,1]$? I am confident the answer is yes however I am having difficulty showing this, here is my attempt: for any $\epsilon>0$ I need to find an $N$, $k \geq N \implies |(1-x)x^k|<\epsilon$ for all $x \in [0,1]$.

Attempt:

Here is my attempt to choose an $N$ independent of $x$.

Since the function is continuous on a closed interval there is a $c \in (0,1)$ such that $|(1-x)x^k|\leq |(1-c)c^k|$ for all $x \in [0,1]$. Then $|(1-c)c^k|<\epsilon \implies c^k< \frac{\epsilon}{1-c}$. Now I am stuck, how am I supposed to choose $N$ here?

Answer 1

To bound $f_k$, note that for $x\ge 1-\frac1{\sqrt k}$, $$\tag1 f_k(x)\le 1-x=\frac1{\sqrt k}$$ and for $x\le 1-\frac1{\sqrt k}$, $$\tag2f_k(x)\le x^k\le\left(1-\frac1{\sqrt k}\right)^k\approx e^{-\sqrt k}$$ so that $\sup f_k\to 0$ as desired.

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To make $(2)$ more explicit, note that for $k\ge 2$ by Bernoulli's ineqaulity, $$\frac1{\left(1-\frac1{\sqrt k}\right)^k} =\left(1+\frac1{\sqrt k-1}\right)^k \ge 1+\frac k{\sqrt k-1}>\sqrt k$$ so that $$ f_k(x)<\frac1{\sqrt k}$$

Answer 2

Call your sequence $f_k(x)$. Consider first finding the maximum as a function of $k$. The derivative gives:

$$-x^k+(1-x)kx^{k-1}=0$$

and when $x\neq 0$, $(1-x)k = x$, giving $x=k/(1+k)$. Using a second derivative test, show that this is indeed a maximum. So $f_k(x)\leq f_k(k/(1+k))$.

$$f_k((k/(1+k))=(1/(k+1))(k/(1+k))^k\rightarrow 0$$

which should give you uniform convergence in $x$.

Answer 3

Set $f_k(x)=(1-x)x^k$. Note that $f_k$ is positive and since $f_k$ is continuous on the compact interval, it admits its maximum value. We can solve $\frac{d}{dx}f_k(x)=0$ to find the maximum. We have that $\frac{df_k}{dx}(x)=kx^{k-1}-(k+1)x^k$, so $kx^{k-1}-(k+1)x^k=0$ if and only if $x=0$ or $x=\frac{k}{k+1}$. It is easily verified that the maximum value of $f_k$ occurs at $x_k=\frac{k}{k+1}$.

So $$\sup_{x\in[0,1]}f_k(x)=f_k(x_k)=f_k(\frac{k}{k+1})=\frac{1}{k}\cdot\frac{k}{k+1}\cdot\big(\frac{k}{k+1}\big)^k=\frac{1}{k}\cdot\big(1-\frac{1}{k+1}\big)^{k+1}\to0$$

because $1/k\to0$ while $(1-\frac{1}{k+1})^{k+1}\to e^{-1}$. Since the supremum of $f_k$ tends to $0$ as $k$ becomes large, we have that $f_k$ converges to $0$ uniformly.

Answer 4

For each $n\in\mathbb{N}$, define $f_{n}:[0,1]\rightarrow(1-x)x^{n}$. We go to show that $f_{n}\rightarrow0$ uniformly.

Proof: Let $\varepsilon>0$ be given. Choose $\delta\in(0,\min(\varepsilon,\frac{1}{2})).$ Since $(1-\delta)^{n}\rightarrow0$, there exists $N$ such that $(1-\delta)^{n}<\varepsilon$ whenever $n\geq N$. Let $n\geq N$ and $x\in[0,1]$. We assert that $|f_{n}(x)-0|<\varepsilon$. If $x\in[0,1-\delta]$, then $0\leq f_{n}(x)\leq x^{n}\leq(1-\delta)^{n}<\varepsilon$. Hence, $|f_{n}(x)-0|<\varepsilon$. If $x\in(1-\delta,1]$, then $0\leq f_{n}(x)\leq1-x<\delta\leq\varepsilon$. We also have $|f_{n}(x)-0|<\varepsilon$. Therefore $f_{n}\rightarrow0$ uniformly.