Analysis Problem of the Day 78

Today’s problem appeared as Problem 3 on the UCLA Fall 2009 Analysis Qual:

Problem 3. Let $f: [0,1] \to \mathbb{R}$ be continuous with $\min_{x \in [0,1]} f(x) =0,$ and suppose that for all $a \leq b,$

$\int_a^b f(x)-(\min_{t \in [a,b]} f(t)) dx \leq \frac12 |b-a|.$

a) Show that for all $\lambda \geq 0,$ $|\{x:f(x)>\lambda+1\}| \leq \frac12 |\{x:f(x)>\lambda\}|.$

b) Prove that for $1 \leq c < 2,$

$\int_0^1 c^{f(x)} dx \leq \frac{100}{2-c}.$

Solution: a) Multiplying by 2 on both sides and using $|\{x: f(x)>\lambda\}|-|\{x:f(x)>\lambda+1\}|=|\{x: \lambda < f(x) \leq \lambda+1\}|,$ we note that it suffices to prove

$|\{x:f(x) > \lambda+1\}| \leq |\{x: \lambda < f(x) \leq \lambda+1\}|$

for all $\lambda \geq 0.$ Since $f$ is continuous, note that $A_\lambda:=\{x:f(x)>\lambda\}$ is open for all $\lambda>0.$ Now, take any interval $[a,b] \subset A_\lambda$ and apply the given condition to obtain

$\int_a^b f(x)-\lambda dx \leq \int_a^b f(x) - \min_{[a,b]} f(t) dx \leq \frac12 |b-a|,$

i.e. $\frac{1}{b-a} \int_a^b f(x) dx \leq \lambda+\frac12.$ Thus, we conclude that the averages of $f$ on intervals in $A_\lambda$ are at most $\frac12$ above $\lambda.$ By the definition of the average, this implies that $\int_{f>\lambda+\frac12} f(x) dx \leq \int_{f \leq \lambda+\frac12} f(x)dx,$ so that $\int_{f >\lambda+1} f(x) dx \leq \int_{\lambda \leq f \leq \lambda+1} f(x)dx.$ We now apply Chebyshev’s inequality to obtain the desired result. Indeed, since $f$ is non-negative,

$|\{x:f(x)>\lambda+1\}| \leq \frac{\int_{f>\lambda+1} f(x)dx}{\lambda+1} \leq \frac{\int_{\lambda \leq f < \lambda+1} f(x)dx}{\lambda+1} \leq |\{x: \lambda \leq f < \lambda+1\}|,$

as desired.

b) We multiply by the denominator on both sides and apply the layer-cake decomposition to obtain

$\int_0^\infty 2|\{x:c^{f(x)}>\lambda\}|-|\{x:c^{f(x)+1}>\lambda\}| dx=\int_0^\infty 2\left|\left\{x:f(x)>\frac{\log \lambda}{\log c}\right\}\right|-\left|\left\{x:f(x)>\frac{\log \lambda}{\log c}-1\right\}\right| dx.$

Now, for $\widetilde{\lambda}:=\frac{\log \lambda}{\log c} \geq 1,$ the claim in (a) gives us that the integrand is bounded above by 0. But note that this holds only if $\log \lambda \geq \log c,$ i.e. $\lambda \geq c.$ We can thus conclude that

$\int_0^1 c^{f(x)}dx \leq \int_0^c 2|\{x:c^{f(x)}>\lambda\}|-|\{x:c^{f(x)}>\lambda+1\}| d\lambda \leq 4.$

Thus, $\int_0^1 c^{f(x)} dx \leq \frac{4}{2-c} \leq \frac{100}{2-c},$ as claimed.

Leave a Reply Cancel reply