fred_91 said:
Thank you.
I see, so I have:
|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(y)(x-y)dy|\le M
Using the triangle inequality, we have:
|\frac{1}{2}f(a)(2xa+a^2)|+|\int_0^a f(y)(x-y)dy|\le M
The first term is bounded by
\frac{1}{2}f(a)(2ba+a^2)
So, we can write:
|\frac{1}{2}f(a)(2ab+a^2)|+\int_0^a |f(y)(x-y)|dy\le M
Well, you shouldn't write ##\leq M## yet because you haven't shown that. Instead, you can write
$$\begin{align}
\left|\frac{1}{2}f(a)(2xa + a^2) - \int_{0}^{a} f(y)(x-y)dy \right|
&\leq \left| \frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a}f(y)(x-y)dy \right|\\
\end{align}$$
Now, you said the first term is bounded by ##\frac{1}{2}f(a)(2ba + a^2)##, but even after adding the missing absolute value signs, that isn't necessarily true if ##a## and ##b## are allowed to be negative. Consider for example ##a=-1##, ##b=0##. However, you can get a bound by applying the triangle inequality again:
$$\begin{align}
\left| \frac{1}{2}f(a)(2xa + a^2)\right|
&\leq \frac{1}{2}|f(a)|(|2xa| + |a^2|)\\
&= \frac{1}{2}|f(a)|(2|a|\cdot|x| + |a|^2)\\
\end{align}$$
Now find a bound for ##|x|## and that takes care of the first term.
For the second term, you need to use the fact that ##f## is continuous, hence bounded on ##[0,a]## (why?), i.e. there exists some bound ##B## such that ##|f(y)| \leq B## for all ##y \in [0,a]##. Now, can you find a bound for ##|x-y|##? Remember that ##y \in [0,a]## and ##x \in [a,b]##.