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Let $f_n(x)=x^{\frac{1}{2n+1}}$, prove wether or not $f_n(x)$ converges uniformly towards a function $f$. Where $x \geq 0$.

First i found looked at pointwise convergence, and i found: \begin{align*} f(x)= \begin{cases} 1& \text{if $x>0$}\\ 0& \text{hvis $x=0$} \end{cases} \end{align*} I am in no way certain that this is correct.

But how would i go about proving wether or not this converge uniformly, if it is correct. Since it is a piecewise function im at a total loss. Normally i would just subtract the function, maximize and show that it went towards zero.

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The ellya's answer is ample but since in your question you suggest to use the definition so let's do it. We have

$$\sup_{x\ge0}|f_n(x)-f(x)|\ge \left|f_n\left(\frac1n\right)-f\left(\frac1n\right)\right|=\left(\frac1n\right)^{\frac1{2n+1}}\xrightarrow{n\to\infty}1\ne0$$ hence we have the desired result.

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It does not converge uniformly since the pointwise limit that you have found (Which is correct by the way), is not continuous.

If it was uniform, then $\lim_{n\to\infty}\sup_x|f_n(x)-f(x)|=0$

But $\sup_x|f_n(x)-f(x)|\ge|f_n(\frac{1}{2})-f(\frac{1}{2})|=|(\frac{1}{2})^{\frac{1}{2n+1}}-1|\to 1 $ as $n\to\infty$

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Hint: uniform convergence preserves continuity.

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