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I want to show that $C^1[0,1]$ isn't a Banach Space with the norm:

$$||f||=\max\limits_{y\in[0,1]}|f(y)|$$

Therefore, I want to show that the sequence $\left \{ |x-\frac{1}{2}|^{1+\frac{1}{n}} \right \}$ converges to $|x-\frac{1}{2}|$, but I can't find $N$ in the definition of convergence. Could anyone give me an idea?

Thank You.

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    You can use that $\lim_n (a_n ^{b_n})=(\lim_n a_n) ^{(\lim_n b_n)}$ – 2017-02-12
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    But, is it always true?, Are there assumptions? – 2017-02-12
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    Just continuity – 2017-02-12

2 Answers 2

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You have, using the Mean Value Theorem, the inequality $$ 1-e^t\leq |t|,\ \ \ \ t\leq0. $$ Then, for $t\in[0,1]$ and $x>0$, $$ |t^{1+x}-t|=|t|\,|t^x-1|\leq |t|\,|e^{x\log t}-1|\leq x\,|t\log t|\leq \frac xe. $$ So $$ \left|\,\left|x-\tfrac12\right|^{1+\tfrac1n}-\left| x-\tfrac12\right|\,\right |\leq\frac1{ne} $$

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    You are welcome. Consider upvoting when you have enough rep, and I hope you keep participating on the site. – 2017-02-12
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To show $\left|x-\frac12\right|^{1+1/n}=\left|x-\frac12\right|\left|x-\frac12\right|^{1/n}\to \left|x-\frac12\right|$ pointwise, which is clear when $x=\frac12$, it suffices to show that when $00$. The subsequence $(t^{1/(2n)})$ must converge to the same limit $L$, and the sequences of squares of this subsequence must converge to $L^2$, implying that $L^2=L$. Hence $L=1$.

From the above, we have monotone pointwise convergence of the sequence to $\left|x-\frac12\right|$. By Dini's Theorem, the convergence is uniform, hence convergence in the given norm.