Puntual and uniform convergence of function

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.

• 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 $\epsilon$. – logarithm May 8 at 13:10

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.