A proof in Hilbert & Courant vol 1 of Weierstrass theorem.

Question

My question is regarding a derivation of an inequality on page 67 of Methods of Mathematical Physics.

Here's a scan of the book: http://web.student.chalmers.se/~robiand/home/files/0.resources/Hilbert-Methods_of_mathematical_physics.pdf

I don't understand how did they get the inequality: $$ | P_n(x) - f(x)|< 2M J_n^*/J_n +\epsilon$$

If I follow the inequalities on this page and the previous integrals on the previous page, I get the following inequality:

$$|P_n(x)-f(x)| < M \frac{J_n^*}{J_n} + \epsilon + |2(J_n-J_n^*)-1|M/(2J_n)$$

Where from $|I_1| , |I_3| < MJ_n^*$ we get the first term, the second term is from the second term in $I_2$ , and the third term is from the first term in $I_2$ and reducing from it $f(x)$ and taking absolute value.

What am I missing here?

Answer 1

That would have been nice for the author to show a little more work there. :-)

The ingredients are:

1. We define $I_1$, $I_2$, and $I_3$ so that $P_n(x) = \frac{I_1 + I_2 + I_3}{2J_n}$.
2. $|I_1|$ and $|I_3|$ are smaller (in modulus) than $M{J_n}^*$.
3. $|I_2|$ is $2f(x)(J_n-{J_n}^*)+ E$, where $|E|<2\epsilon J_n$.
4. $|f(x)| <M$.

Therefore $$|P_n(x) - f(x)| = \left|\frac{I_1 + I_2 + I_3}{2J_n} - f(x)\right| = \left|\frac{I_1 + I_2 + I_3 - f(x) \cdot 2 J_n}{2J_n}\right|$$ $$\leq \left|\frac{I_2 - f(x) \cdot 2 J_n}{2J_n}\right| + \left| \frac{I_1 + I_3}{2J_n} \right| = \left|\frac{2f(x)(J_n - {J_n}^*) + E - f(x) \cdot 2 J_n}{2J_n}\right| + \left| \frac{I_1 + I_3}{2J_n} \right|$$ $$= \left|\frac{E-2f(x){J_n}^*}{2J_n}\right| + \left| \frac{I_1 + I_3}{2J_n} \right| \leq \left| \frac{E}{2J_n}\right| + \left| \frac{2f(x){J_n}^*}{2J_n}\right| + \left| \frac{I_1 + I_3}{2J_n}\right|$$ $$ \leq \epsilon + M\left| \frac{{J_n}^*}{J_n} \right| + \left| 2\frac{M{J_n}^*}{2J_n} \right|$$ $$= \epsilon + M \left| \frac{{J_n}^*}{J_n} \right| + M\left| \frac{{J_n}^*}{J_n} \right| = \epsilon + 2M\frac{{J_n}^*}{J_n}.$$