Is this sequence of functions uniformly convergent on $[0,1]$?

Solution 1:

As $c$ is a positive integer, so for $t > 0$ and for all $x \in [0, 1]$, we have \begin{align} \left\lvert f_t(x) - f(x) \right\rvert &= \left\lvert t x^c e^{tx} - 0 \right\rvert \\ &= t x^c e^{tx} \\ &\leq t 1^c e^{t \times 1} \\ & \mbox{[ because the functions $x^c$ and $e^{tx}$ are strictly increasing on $[0, 1]$, } \\ & \qquad \mbox{ a fact you can verify using the signs of the derivatives ] } \\ &= t e^t. \end{align} Therefore we have \begin{align} \lim_{t \to 0+0} \max_{x \in [0, 1] } \left\lvert f_t(x) - f(x) \right\rvert &= \lim_{t \to 0+0} t e^t \\ &= 0 e^0 \\ &= 0, \end{align}

For $t < 0$, let us put $u = -t$ so that $u > 0$. Then for all $x \in [0, 1]$, we have \begin{align} \left\lvert f_t(x) - f(x) \right\rvert &= \left\lvert t x^c e^{tx} - 0 \right\rvert \\ &= -t x^c e^{tx} \\ &= u x^c e^{-ux} \\ &\leq u 1^c e^{-u \times 0} \\ & \mbox{[ because the functions $x^c$ is strictly increasing on $[0, 1]$ } \\ & \qquad \mbox{ and $e^{-ux}$ is strictly decreasing on $[0, 1]$, } \\ & \qquad \mbox{ facts you can verify using the signs of the derivatives ] } \\ &= u. \end{align} Therefore we have \begin{align} \lim_{t \to 0-0} \max_{x \in [0, 1] } \left\lvert f_t(x) - f(x) \right\rvert &= \lim_{u \to 0+0} u \\ &= 0, \end{align} as required.

Hope this helps.