Problem: Let $f\in(\mathcal C[0,1],\|\cdot\|_\infty)$ such that $f(0)=0$. Define for $n\in\mathbb N$ $$(T_nf)(t)=\int_{0}^{t}e^{-n(t-x)}f(x)\,dx\quad\text{for }t\in[0,1].$$ $\textbf{a)}$ Show that $\|T_nf\|_\infty\leq\displaystyle\frac{1-e^{-n}}{n}\|f\|_\infty.$
$\textbf{Proof Attempt:}$ We observe that \begin{equation*}\begin{split} \vert T_nf(t)\vert=\Bigg\vert\int_{0}^{t}e^{-n(t-x)}f(x)\,dx\Bigg\vert &\leq \int_{0}^{t}\vert e^{-n(t-x)}f(x)\vert\,dx\\ &\leq \|f\|_\infty\int_{0}^{t}e^{-n(t-x)}\,dx\\ &= \frac{1-e^{-nt}}{n}\|f\|_\infty\\ &\leq \frac{1-e^{-n}}{n}\|f\|_\infty. \end{split}\end{equation*}
$\textbf{b)}$ Show that $$nT_nf(t)-f(t)=\int_{0}^{t}ne^{-nx}[f(t-x)-f(t)]\,dx+e^{-nt}f(t)$$ for all $t\in[0,1]$ and $n\in\mathbb N.$
$\textbf{Proof Attempt:}$ Since $$\int_{0}^{t}e^{-nx}f(t-x)\,dx=\int_{0}^{t}e^{-n(t-x)}f(x)\,dx\quad\text{and}\quad \int_{0}^{t}e^{-nx}\,dx=\frac{1-e^{nt}}{n},$$ we see that \begin{equation*}\begin{split} &\quad\int_{0}^{t}ne^{-nx}[f(t-x)-f(t)]\,dx+e^{-nt}f(t)\\ &=n\int_{0}^{t}e^{-nx}f(t-x)\,dx-nf(t)\int_{0}^{t}e^{-nx}\,dx+e^{-nt}f(t)\\ &= nT_nf(t)-nf(t)\frac{1-e^{nt}}{n}+e^{-nt}f(t)\\ &= nT_nf(t)-f(t). \end{split}\end{equation*}
$\textbf{c)}$ Show that $\lim\limits_{n\to\infty}nT_nf=f$ uniformly on $[0,1]$.
$\textbf{Proof Attemtp:}$ Here I will use the following Lemma: If $h\in\mathcal B(0,1]$, that is, it is bounded on $(0,1]$, such that $\lim\limits_{t\searrow0}\,h(t)=0$, then for $g_n(t)=e^{-nt}h(t)$ where $t\geq0$, we have that $g_n\to0$ uniformly on $(0,1].$ Using the Lemma we have that $e^{-nt}f(t)\to0$ uniformly on $[0,1]$ since also$f(0)=0$ by hypothesis. It remains to handle the integral.
Let $\varepsilon>0$ be given. Since $\frac{n}{e^{nx}}\to0$ uniformly on $[\delta,1]$, there exists some $N\in\mathbb N$ such that $\frac{n}{e^{nx}}\leq\varepsilon$ for all $n>N$ and all $x\in[\delta,1]$ where $1>\delta>0$ is fixed. We also choose $\delta$ such that $\vert f(t-x)-f(t)\vert\leq\varepsilon$ for all $\vert x\vert\leq\delta$ by uniform continuity. Then for all $n>N$ and all $t\in[0,1]$ we have that \begin{align*} \Bigg\vert\int_{0}^{t}ne^{-nx}[f(t-x)-f(t)]\,dx\Bigg\vert &\leq \varepsilon\int_{0}^{\delta}ne^{-nx}\,dx+2\|f\|_\infty\int_{\delta}^{1}ne^{-nx}\,dx\\ &\leq\varepsilon(1-e^{-n\delta})+2\|f\|_\infty\varepsilon(1-\delta)\\&\leq\varepsilon+2\|f\|_\infty\varepsilon. \end{align*} Therefore, $\displaystyle\int_{0}^{t}ne^{-nx}[f(t-x)-f(t)]\,dx\to0$ uniformly on $[0,1].$
$\textbf{My Concern:}$ I think I was not successful at handling the integral in part (c).
$\color{blue}{\textbf{Could anyone please check if my proof for part (c) is correct}}?$
Thank you for your time and appreciate any feedback. Comments on parts (a) and (b) are also welcomed.
Your proof in part (c) looks good. Maybe it would be better to add a step where you compute the integral $\int_{0}^{\delta}ne^{-nx}\,dx$ explicitely in order to make the bound more visible.