Suppose $$ -u''(x) = f(x) $$ on $(0,1)$, and $u(0)=u(1)=0$, where $u$ is the unknown and $f$ is a continuous smooth term.
How to show that $$ \|u\|_ {\infty} \le \frac{1}{8}\,\|f\|_{\infty} ? $$
I was looking for Taylor's formula with integral remainder
you have
$$ -u''(x) = f(x)\implies -u'(t) = \int_0^tf(s)ds +c \implies -u(x) = \int_0^x \left(\int_0^tf(s)ds +c\right)dt+ a$$
$u(0)= 0$ then $a=0$ and $$u(1)= \int_0^1 \left(\int_0^tf(s)ds +c\right)dt = \int_0^x \left(\int_0^tf(s)ds +c\right)dt =0 \\\implies c= -\int_0^1 \left(\int_0^tf(s)ds \right)dt $$ This implies $$ -u(x) = -x\int_0^1 \left(\int_0^tf(s)ds \right) +\int_0^x \left(\int_0^tf(s)ds \right)dt $$
$$|u(x)|\le \left(x\int_0^1 tdt +\int_0^x tdt\right)\|f\|_{\infty}$$