Let $f$ be a non constant function defined on $[a, b]$ and $f(a)=f(b)=0$. Prove that there exists a $\xi \in (a, b)$ such that:
$$\bigl|{f'(\xi)}\bigr|>\frac{4}{(b-a)^2}\int_{a}^{b}{|f(x)|\,{\rm d}x}$$
Solution
To simplify the notation we define \(g:[0,1] \to \mathbb{R}\) by \(\displaystyle{g(x) = f\left( (b-a)x + a\right)}.\) Trivially, \(g\) is non-zero differentiable function with \(g(0) = g(1) = 0.\) We claim that it is enough to find \(\vartheta \in [0,1]\) such that \[ |g'(\vartheta)| > 4\int_0^1 |g(x)| \, dx.\] Indeed then we have \[ |f'((b-a)\vartheta + a)| = \frac{|g'(\vartheta)|}{b-a} > \frac{4}{b-a}\int_0^1 g(x) \, dx = \frac{4}{(b-a)^2}\int_a^b f(t) \, dt,\] where in the last equality we used the substitution \(t=(b-a)x+a.\)
Now let \(M = 4\int_0^1 |g(x)| \, dx.\) We may assume that \(M > 0\) and that \(|g'(x)| \leqslant M\) for each \(x \in [0,1]\). We claim that \(g(x) \leqslant Mx\) for each \(x \in [0,1]\). Indeed if \(g(y) > My\) for some \(y\) then by the mean value theorem there would be a \(x \in (0,y)\) with \[g'(x) = \frac{g(y) - g(0)}{y-0}> M,\] a contradiction. Similarly it holds that \(g(x) \leqslant M(1-x)\) for each \(x \in [0,1].\) But then \[ \frac{M}{4} = \int_0^1 g(x) \, dx \leqslant \int_0^{1/2} Mx \, dx + \int_{1/2}^1 M(1-x) \, dx = \frac{M}{4}.\] So we must have equality and because \(g\) is continuous it must be equal to \(Mx\) in \([0,1/2]\) and equal to \(M(1-x)\) in \([1/2,1]\). But then \(g\) is not differentiable at \(1/2\) a contradiction.
This exercise has also been discussed in mathimatikoi.org long before it was published in the exams.
$$\bigl|{f'(\xi)}\bigr|>\frac{4}{(b-a)^2}\int_{a}^{b}{|f(x)|\,{\rm d}x}$$
(Qualifying Exams, Wisconsin-Madison, 2015)
Solution
To simplify the notation we define \(g:[0,1] \to \mathbb{R}\) by \(\displaystyle{g(x) = f\left( (b-a)x + a\right)}.\) Trivially, \(g\) is non-zero differentiable function with \(g(0) = g(1) = 0.\) We claim that it is enough to find \(\vartheta \in [0,1]\) such that \[ |g'(\vartheta)| > 4\int_0^1 |g(x)| \, dx.\] Indeed then we have \[ |f'((b-a)\vartheta + a)| = \frac{|g'(\vartheta)|}{b-a} > \frac{4}{b-a}\int_0^1 g(x) \, dx = \frac{4}{(b-a)^2}\int_a^b f(t) \, dt,\] where in the last equality we used the substitution \(t=(b-a)x+a.\)
Now let \(M = 4\int_0^1 |g(x)| \, dx.\) We may assume that \(M > 0\) and that \(|g'(x)| \leqslant M\) for each \(x \in [0,1]\). We claim that \(g(x) \leqslant Mx\) for each \(x \in [0,1]\). Indeed if \(g(y) > My\) for some \(y\) then by the mean value theorem there would be a \(x \in (0,y)\) with \[g'(x) = \frac{g(y) - g(0)}{y-0}> M,\] a contradiction. Similarly it holds that \(g(x) \leqslant M(1-x)\) for each \(x \in [0,1].\) But then \[ \frac{M}{4} = \int_0^1 g(x) \, dx \leqslant \int_0^{1/2} Mx \, dx + \int_{1/2}^1 M(1-x) \, dx = \frac{M}{4}.\] So we must have equality and because \(g\) is continuous it must be equal to \(Mx\) in \([0,1/2]\) and equal to \(M(1-x)\) in \([1/2,1]\). But then \(g\) is not differentiable at \(1/2\) a contradiction.
This exercise has also been discussed in mathimatikoi.org long before it was published in the exams.
No comments:
Post a Comment