Jtdylktuy

Let $f:[0,1] to mathbb{R}$ a continuous function such $f(0)=f(frac{1}{2})=0$ and $(f[0, frac{1}{2}]) not subset lbrace 0 rbrace$. For each $n in mathbb{N}$, let $g_{n}:[0, infty ) to mathbb{R}$ defined for each $x in [0, infty)$ as $$g_{n}(x)=fleft(frac{sqrt[n]{x}}{1+ sqrt[n]{x}}right)$$
Prove the following:

(a) $lbrace g_{n} rbrace_{n=1}^{infty}$ is pointwise convergent to $0$.

(b) $lbrace g_{n} rbrace_{n=1}^{infty}$ is not uniformly continuous to $0$.

My attemp for (a) goes as follow: As

$$left| frac{sqrt[n]{x}}{1+ sqrt[n]{x}} right|= left| left( sqrt[n]{x}right) frac{1}{1+ sqrt[n]{x}} right| = left| sqrt[n]{x} right| left| frac{1}{1+ sqrt[n]{x}} right|$$. I want to use the fact that as $sqrt[n]{x} to 1$ as $n to infty$ for $x in [0,1]$ and $frac{1}{1+ sqrt[n]{x}} to frac{1}{2}$ as $n to infty$ for $x in [0,1]$ then we can bound

$$left| frac{sqrt[n]{x}}{1+ sqrt[n]{x}} - frac{1}{2} right|< epsilon.$$

For every $epsilon>0$ and of course a particular $ngeq N(epsilon) in mathbb{N}$. Then as $f$ is continuous in [0,1] then

$$ g_{n}(x)=fleft(frac{sqrt[n]{x}}{1+ sqrt[n]{x}}right) to fleft(frac{1}{2}right)=0$$

Im kind of lost in proving the not uniformly continuity of $g_{n} to 0$. Any help ending the proof would be appreciated. Thanks

For (a) you need to distinguish between $x=0$ and $x>0.$ If $x=0,$ then $g_n(x) = f(0)=0$ for all $n.$ If $x>0,$ then $x^{1/n} to 1,$ hence $x^{1/n}/(1+x^{1/n}) to 1/2,$ which implies $g_n(x) to f(1/2)=0.$

For (b), we know that there is $cin (0,1/2)$ such that $f(c)ne 0.$ This $c$ can be written as $x/(1+x)$ for some $xge 0.$ We then have $g_n(x^n) = f(x/(1+x))=f(c)$ for all $n.$ Thus

$$sup_{mathbb [0,infty)}|g_n| ge |g_n(x^n)| = |f(c)| >0$$

for all $n.$ This shows $g_n$ does not converge uniformly to $0$ on $[0,infty).$

We show that the function is not uniformly continuous even over $(1,infty)$. First notice that since $f(x)$ is continuous over $[0,1]$ we have:$$0<x<delta_epsilonquad,quad |f(x)|<epsilon $$therefore $$forall n>Nquadto quad 0<{sqrt[n]xover 1+sqrt[n]x}<delta_epsilonto |g_n(x)|=|f({sqrt[n]xover 1+sqrt[n]x})|<epsilon$$also $0<{sqrt[n]xover 1+sqrt[n]x}<delta_epsilon$ is equivalent to $$n>left({1over delta_epsilon}-1right){1over ln x}=N$$. Uniform continuity requires that $N=left({1over delta_epsilon}-1right){1over ln x}$ be a function only of $epsilon$ while we see that a term $1overln x$ makes it dependent also to $x$. Therefore $g_n(x)$ tends to $0$ though not uniformly.