Bourbaki marks this one as harder than the rest.
a) Let f be a twice differentiable real function on R, such that (f(x))2≤a and (f′(x))2+(f′′(x))2≤b on R; show that
(f(x))2+(f′(x))2≤max(a,b) on
R (argue by contradiction, noting that if the function
f2+f′2 takes a value
c>max(a,b) at a point
x0 then there exist two points
x1,x2 such that
x1<x0<x2 and that at
x1 and
x2 the function
f′ takes values small enough that
f2+f′2 takes values
<c; then consider a point of
[x1,x2] where
f2+f′2 attains its supremum on this interval).
b) Let f be a real function n times differentiable on R, and such that (f(x))2≤a and (f(n−1)(x))2+(f(n)(x))2≤b on R; show that then
(f(k−1)(x))2+(f(k)(x))2≤max(a,b) on
R for
1≤k≤n. (Argue by induction on
n; note that, by exerc. 12 the supremum
c of
(f′(x))2 on
R is finite; show that one necessarily has
c≤max(a,b) by reducing to a contradiction: assuming that
c>max(a,b) choose the constants
λ and
μ so that for the function
g=λf+μ one has
∣g(x)∣≤1, ∣g′(x)∣≤1, yet one cannot have
(g(x))2+(g′(x))2≤1 for all
x.)