11.7. EXERCISES 293

Conclude that if s+ 1−2t2 = s+ 1

2 − t ≤ 0 and t > 1/2, then ∥γ(u)∥2s,2 ≤Cn ∥u∥2

t,2. Inparticular, if t = 1 and s = 1/2, you have ∥γ(u)∥2

(1/2),2 ≤Cn ∥u∥21,2 . This exhibits the

phenomenon of the loss of 1/2 derivative when considering γu on an n− 1 dimen-sional subspace.

17. In dealing with Sobolev spaces, the following interpolation inequality is very useful.Let 0≤ r < s < t. Then if u ∈ Ht (Rn) ,

∥u∥Hs(Rn) ≤ ∥u∥θ

Hr(Rn) ∥u∥1−θ

Ht (Rn)

where θ ∈ (0,1) such that θr+(1−θ) t = s. Hint:

∥u∥Hs(Rn) =

(∫Rn

(1+ |x|2

)θr(1+ |x|2

)(1−θ)t|Fu(x)|2 dx

)1/2

.

Regard |Fu(x)|2 dx = dµ as a measure and use Holder’s inequality.

(∫Rn|Fu(x)|2

(1+ |x|2

)sdx)1/2

=

(∫Rn

(1+ |x|2

)θr(1+ |x|2

)(1−θ)t|Fu(x)|2 dx

)1/2

≤

(∫Rn

(1+ |x|2

)r|Fu(x)|2 dx

)θ

·(∫Rn

(1+ |x|2

)t|Fu(x)|2 dx

)1−θ

1/2

= ∥u∥θ

Hr(Rn) ∥u∥1−θ

Ht (Rn)

18. If ε > 0 and if θr+(1−θ) t = s,0≤ r < s < t, show that there exists a constant Cε

such that∥u∥Hs(Rn) ≤ ε ∥u∥Ht (Rn)+Cε ∥u∥Hr(Rn)

Hint: For r large and positive,

∥u∥Hs(Rn) ≤(

r∥u∥Hr(Rn)

)θ 1rθ∥u∥1−θ

Ht (Rn)

=(

r∥u∥Hr(Rn)

)θ

((1rθ

)1/(1−θ)

∥u∥Ht (Rn)

)1−θ

.

Now recall Proposition 9.3.2 and then pick r sufficiently large. This is very use-ful for checking conditions needed in non-linear partial differential equations andinclusions.