1. **Problem statement:** We analyze the convergence (pointwise and uniform) of two sequences of functions: (a) $f_n(x) = \frac{n x^2}{e^{n x^2}}$ on $\mathbb{R}$. (b) $f_n(x) = n^c x (1 - x^2)^n$ on $(0,1)$ for $c \in \mathbb{R}$. 2. **Sequence (a):** $f_n(x) = \frac{n x^2}{e^{n x^2}}$ - For fixed $x$, consider the limit as $n \to \infty$. - If $x = 0$, then $f_n(0) = 0$ for all $n$. - If $x \neq 0$, write $f_n(x) = n x^2 e^{-n x^2}$. - Since $e^{n x^2}$ grows faster than any polynomial in $n$, $f_n(x) \to 0$. 3. **Pointwise limit:** $$ \lim_{n \to \infty} f_n(x) = 0 \quad \text{for all } x \in \mathbb{R}. $$ 4. **Uniform convergence:** - To check uniform convergence, consider the supremum norm: $$ \|f_n\|_\infty = \sup_{x \in \mathbb{R}} |f_n(x)|. $$ - Maximize $f_n(x)$ over $x$. - Set derivative w.r.t. $x$ to zero: $$ \frac{d}{dx} \left( n x^2 e^{-n x^2} \right) = 0. $$ - Compute derivative: $$ 2 n x e^{-n x^2} - 2 n^2 x^3 e^{-n x^2} = 2 n x e^{-n x^2} (1 - n x^2) = 0. $$ - Critical points at $x=0$ and $x = \pm \frac{1}{\sqrt{n}}$. - Evaluate $f_n$ at $x = \frac{1}{\sqrt{n}}$: $$ f_n\left( \frac{1}{\sqrt{n}} \right) = n \left( \frac{1}{\sqrt{n}} \right)^2 e^{-n \left( \frac{1}{\sqrt{n}} \right)^2} = n \frac{1}{n} e^{-1} = e^{-1}. $$ - Since $\|f_n\|_\infty = e^{-1}$ for all $n$, the supremum does not go to zero. 5. **Conclusion for (a):** - Pointwise limit is zero. - Not uniformly convergent on $\mathbb{R}$ because $\|f_n\|_\infty \not\to 0$. 6. **Sequence (b):** $f_n(x) = n^c x (1 - x^2)^n$ on $(0,1)$. - For fixed $n$, find $x_n \in (0,1)$ maximizing $f_n(x)$. - Take derivative w.r.t. $x$: $$ \frac{d}{dx} f_n(x) = n^c \left( (1 - x^2)^n - 2 n x^2 (1 - x^2)^{n-1} \right) = 0. $$ - Simplify: $$ (1 - x^2) - 2 n x^2 = 0 \implies 1 - x^2 = 2 n x^2 \implies 1 = x^2 (1 + 2 n) \implies x_n = \frac{1}{\sqrt{1 + 2 n}}. $$ 7. **Evaluate $f_n(x_n)$:** $$ f_n(x_n) = n^c \frac{1}{\sqrt{1 + 2 n}} \left( 1 - \frac{1}{1 + 2 n} \right)^n = n^c \frac{1}{\sqrt{1 + 2 n}} \left( \frac{2 n}{1 + 2 n} \right)^n. $$ - Rewrite: $$ \left( \frac{2 n}{1 + 2 n} \right)^n = \left( 1 - \frac{1}{1 + 2 n} \right)^n. $$ - As $n \to \infty$, $\left( 1 - \frac{1}{1 + 2 n} \right)^n \to e^{-\frac{1}{2}}$. 8. **Asymptotic behavior:** $$ f_n(x_n) \sim n^c \frac{1}{\sqrt{2 n}} e^{-\frac{1}{2}} = e^{-\frac{1}{2}} \frac{n^c}{\sqrt{2 n}} = e^{-\frac{1}{2}} \frac{n^{c}}{n^{1/2} \sqrt{2}} = \frac{e^{-1/2}}{\sqrt{2}} n^{c - \frac{1}{2}}. $$ 9. **Limit of $f_n(x_n)$ as $n \to \infty$ depends on $c$:** - If $c < \frac{1}{2}$, then $n^{c - 1/2} \to 0$, so $f_n(x_n) \to 0$. - If $c = \frac{1}{2}$, then $f_n(x_n) \to \frac{e^{-1/2}}{\sqrt{2}}$ (nonzero constant). - If $c > \frac{1}{2}$, then $f_n(x_n) \to \infty$. 10. **Pointwise limit for fixed $x$:** - For fixed $x \in (0,1)$, $(1 - x^2)^n \to 0$ exponentially. - So $f_n(x) = n^c x (1 - x^2)^n \to 0$ for all $c$. 11. **Uniform convergence:** - Since $f_n(x_n)$ is the maximum value, uniform convergence to zero requires $f_n(x_n) \to 0$. - This happens if and only if $c < \frac{1}{2}$. 12. **Summary:** - (a) $f_n$ converges pointwise to zero but not uniformly on $\mathbb{R}$. - (b) $f_n$ converges pointwise to zero on $(0,1)$ for all $c$. - Uniform convergence on $(0,1)$ holds if and only if $c < \frac{1}{2}$.