Proof for an integral inequality

Question

For a function $f:\mathbb{R}\to\mathbb{R}$, $|f|\leq M$, $f(s)\equiv 0$ for $s<0$, define \begin{equation} I^n(t)= n \int_{t-1/n}^{t} f(s)ds. \end{equation} In my book they conclude that \begin{equation}\label{eq} |I^n(t')-I^n(t)|\leq 2n|t'-t|M. \end{equation}

How can I proof this inequality? I only see that \begin{equation} |I^n(t')-I^n(t)|\leq |n[(t'-t'+1/n)M-(t-t+1/n)M]|=0. \end{equation}

Answer 1

Suppose $I^n(t)=\int_{t-1/n}^{t} f(s) \,ds$ and $I^n(t')=\int_{t'-1/n}^{t'} f(s)\,ds$.

Then \begin{split}|I^n(t) - I^n(t')| &= n|\int_{t-\frac{1}{n}}^{t} f(s) \,ds - \int_{t'-\frac{1}{n}}^{t'} f(s)\,ds| \\ &\leq n\left(|\int_{t-\frac{1}{n}}^{t} f(s) \,ds| + |\int_{t'-\frac{1}{n}}^{t'} f(s)\,ds| \right) \\ &= n\left( |\int_{t-\frac{1}{n}}^{t} f(s) \,ds| + |\int_{t'}^{t'-\frac{1}{n}} f(s)\,ds| \right) \\&\leq nM(t - (t - \frac{1}{n}) + nM(t' - \frac{1}{n} - t') \\ &=nM(t - t') + nM(t' - t) \\&\leq 2Mn|t-t'| \end{split}

Answer 2

hint

Put $u=t-s $. then $$I^n (t)=\int_0^\frac 1nf (t-u)du $$

and

$$I^n (t)-I^n (t')=\int_0^\frac 1n (f (t-u)-f (t'-u))du $$