I am facing the following problem: Let $H:[0,1] \to R$ a function such thath $H(0) = 0 = H(1)$, H continous in $x = 0$ and $H(x) \neq 0$ for some $x$. Let $H_n:[0,1] \to R$ be such that $H_n (x) = H(x^n)$.

Is ${H_n}$ convergent? Is it puntually convergent?

Since $H$ is continuous in $x = 0$, given $\epsilon_0 > 0$ there exist $\delta$ such that $f(x) < \epsilon_0$ if $x < \delta$. Clearly, for all $x \in (0,1)$ there exists $\eta$ such that $x^n < \delta$ if $n > \eta$. Therefore, $H^n(x) = H(x^n) < \epsilon_0$ if $n > \eta$.

Since $\epsilon_0$ is arbitrarily chosen, ${{H_n}} \to 0$ in $(0, 1)$. Since $H_n(0) = H_n (1) = 0$, $H_n \to 0$ in $[0, 1]$. This is, $H_n$ converges puntually. Is this correct?

In that case, is it uniformly convergent? My guess is that it isn't since it is possible that $lim_{x \to 1^-} H(x) = +\infty$. But I don't know how to formalize this.

  • 1
    $\begingroup$ Pointwise convergence is okay. To show that it is not necessarily uniformly convergent take some $H$ that is unbounded near $1$. For example, $H(x)=x/(1-x)$ for $x\in[0,1)$ and $H(1)=0$. Then the closer the $x$ is to $1$ the larger the $n$ needs to be such that $H_n(x)$ is smaller than $\epsilon$. $\endgroup$ – logarithm May 8 at 13:10

Uniform convergent or not :

there is $x_0 \in (0,1) $ such that $H(x_0) \ne 0.$ Then we have

$ (*) \quad H_n(x_0^{1/n})=H(x_0) \ne 0.$

Since $(H_n)$ converges pointwise to $0$, $(*)$ shows that $(H_n)$ is not uniformly convergent.

  • $\begingroup$ Thank you, I can see it clearly now. $\endgroup$ – Zanzag May 8 at 13:46

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