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Let $f(x)$ be a $C^2$ function on $[0,1]$, split uniformly into intervals of length $h=1 / n$. Let $p$ be the piecewise linear polynomial interpolating $f$ on each interval.

I am trying to prove the following $L^2$ error estimate:

$\sqrt{\int_0^1|f(x)-p(x)|^2 d x} \leq h^2 \sqrt{\int_0^1\left|f^{\prime \prime}(x)\right|^2 d x}$

My attempt at the proof is to first show that the Taylor expansion of $f(x)$ for $x$ close to zero can be written as $$ f(x)=f(0)+f^{\prime}(0) x+\int_0^x f^{\prime \prime}(s)(x-s) d s $$

and then show $$p(x)=f(0)+f^{\prime}(\alpha) x, \alpha \in[0, h]$$

I would use these to results to show $$|f(x)-p(x)| \leq \sqrt{3 h} x\left(\int_0^h\left|f^{\prime \prime}(s)\right|^2 d s\right)^{1 / 2}$$,

and finally show

$$\|f-p\|_{L^2([0,1])} \leq h^2\left\|f^{\prime \prime}\right\|_{L^2([0,1])}$$

Can someone help me with this proof?

Questioner5453
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1 Answers1

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The trick is to use the Taylor formula based in $x$ to get expressions for $f(x_0)$, $f(x_1)$. As per https://math.stackexchange.com/a/610305/115115 you get for the linear interpolation on some interval $[x_0,x_1]$ $$ f(x)=p(x)+\int_{x_0}^{x_1} G(x,s)f''(s)\,ds\tag{GK} $$ with (using notation 8$a\sqcup b=\max(a,b)$, $a\sqcap b=\min(a,b)$) $$ G(x,s)=\dfrac{(x_1-x\sqcup s)(x\sqcap s-x_0)}{x_1-x_0} =\begin{cases} \dfrac{(x_1-x)(s-x_0)}{x_1-x_0}, & s\in [x_0,x]\\[.5em] \dfrac{(x_1-s)(x-x_0)}{x_1-x_0}, & s\in [x,x_1]\\ \end{cases} $$ So via Cauchy-Schwarz $$ |f(x)-p(x)|^2\le\left(\frac{(x_1-x)^2(x-x_0)^2}{3(x_1-x_0)}\right)\|f''\|^2_{L^2[x_0,x_1]}\tag{CS} $$ and this can now be integrated over the interval \begin{align} \|f-p\|^2_{L^2[x_0,x_1]} &\le\|f''\|^2_{L^2[x_0,x_1]} \int_{x_0}^{x_1}\frac{(x_1-x_0)^2(x-x_0)^2-2(x_1-x_0)(x-x_0)^3+(x-x_0)^4}{3(x_1-x_0)} \\ &=\|f''\|^2_{L^2[x_0,x_1]} \left(\frac{\frac13-\frac12+\frac15}{3}\right)(x_1-x_0)^4 \\ &=\frac{(x_1-x_0)^4}{90}\|f''\|^2_{L^2[x_0,x_1]} \end{align}

One could avoid this last integration using the AM-GM inequality $$ \sqrt{(x_1-x)(x-x_0)}\le\frac{(x_1-x)+(x-x_0)}2=\frac{x_1-x_0}2 $$ directly in (CS) to get $$ |f(x)-p(x)|^2\le\frac{(x_1-x_0)^3}{3ยท2^4}\|f''\|^2_{L^2[x_0,x_1]} \\ \implies \|f-p\|^2_{L^2[x_0,x_1]} \le\frac{(x_1-x_0)^4}{48}\|f''\|^2_{L^2[x_0,x_1]} $$

Lutz Lehmann
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