Math 597. Homework 11 Bryden Cais

(1) Let a, b > 1/2 and define

f(x) = \begin{cases}\frac{1}{\sqrt{x}| log x|^{a}} & if 2 ≤ x < ∞ \\ \frac{1}{\sqrt{x}| log x|^{b}} & if 0 < x < \frac{1}{2} \\ 0 & otherwise \end{cases} .

We claim that f ∈ L^{2}(R, μ_{Leb}) but f ∉ L^{p}(R, μ_{Leb}) for any p ≠ 2. Indeed, we have

∫_{R} |f|^{2} dμ = ∫_{0}^{1/2} \frac{dx}{x| log x|^{2b}} + ∫_{2}^{∞} \frac{dx}{x| log x|^{2a}}

= −\frac{1}{1 − 2b}| log x|^{1−2b}]_{0}^{1/2} + \frac{1}{1 − 2a}(log x)^{1−2a}]_{2}^{∞} < ∞

since a, b > \frac{1}{2} so that both terms above tend to 0 as x → 0 and x → ∞ respectively.

Suppose that p < 2. Then we have

∫_{R} |f|^{p} dμ ≥ ∫_{2}^{∞} \frac{dx}{x^{p/2}(log x)^{pa}}

= ∫_{log 2}^{∞} \frac{dt}{e^{(p/2−1)t}t^{pa}} ,

which diverges since p < 2 so the integrand behaves like t^{−pa}e^{αt} for some α > 0 and hence is unbounded as t → ∞.

On the other hand, if p > 2 then we have

∫_{R} |f|^{p} dμ ≥ ∫_{0}^{1/2} \frac{dx}{x^{p/2}| log x|^{pb}}

= ∫_{−∞}^{− log 2} \frac{dt}{e^{(p/2−1)t}|t|^{pb}}

= ∫_{log 2}^{∞} \frac{dt}{e^{(1−p/2)t}t^{pb}} ,

which diverges since p > 2 so the integrand behaves like t^{−pb}e^{αt} for some α > 0 and hence is unbounded as t → ∞.

Thus, f ∈ L^{2} but f ∉ L^{p} for any p ≠ 2.

(2) (a) Define

f(x) = \begin{cases}n^{2}(x − n) + 1 & if x ∈ [n − 1/n^{2}, n] for each integer n ≥ 2 \\ −n^{2}(x − n) + 1 & if x ∈ [n, n + 1/n^{2}] for each integer n ≥ 2 \\ 0 & otherwise \end{cases} .

Observe that f is continuous and that f(n) = 1 for each integer n ≥ 2. Moreover, we have

∫_{R} |f| dμ = ∑_{n≥2} \frac{1}{2} \frac{2}{n^{2}} < ∞

so that f ∈ L^{1}(R, μ_{Leb}). On the other hand, f(x) ↛ 0 as x → ∞ since f(n) = 1 for every

integer n ≥ 2.

(b) Let f be a function on R with continuous derivative f′ and f, f′ ∈ L^{1}(R, μ_{Leb}). Then by the Fundamental Theorem of Calculus we have

f(x) − f(a) = ∫_{a}^{x} f′(t) dt.

Since f′ ∈ L^{1}(R, μ_{Leb}), the limit

lim_{x→∞} ∫_{a}^{x} |f′(t)| dt

exists and is finite. It follows that

lim_{x→∞}(f(x) − f(a)) = lim_{x→∞} ∫_{a}^{x} f′(t) dt

exists and is finite, so that lim_{x→∞} f(x) = c < ∞ exists and is finite. Pick ϵ > 0 and let N_{ϵ} be such that for all x > N_{ϵ} we have |f(x) − c| < ϵ. Then for x > N_{ϵ} we have

∫_{N_{ϵ}}^{x} |f(t)| dt ≥ ∫_{N_{ϵ}}^{x}(|c| − ϵ) dt = (x − N_{ϵ})(|c| − ϵ).

If c ≠ 0 we can find ϵ > 0 so that |c| − ϵ > 0 whence, taking the limit as x → ∞ above, we find that f ∉ L^{1}(R, μ_{Leb}), a contradiction. It follows that c = 0 so that lim_{x→∞} f(x) = 0.

(3) (a) Let ϕ = f_{1}χ_{I_{1}} +⋯+f_{N}χ_{I_{N}} be a step function and set I = ⋃ I_{j} and F = |f_{1}|+⋯+|f_{N}|. Since each I_{j} is a bounded interval, we have μ(I) < ∞. Let the closure of I_{j} be [a_{j}, b_{j}]. We have

|∫_{R} ϕ(x)e^{−ixt} dx| = |∑_{j=1}^{N} f_{j} \frac{e^{−itb_{j}} − e^{−ita_{j}}}{−it} |

|≤ ∑_{j=1}^{N} f_{j} \frac{2}{t}| ≤ \frac{2F}{|t|} ,

since |e^{−ixt}| ≤ 1 for any x, t ∈ R. It follows that |ϕ̂(t)| → 0 as t → ∞ or −∞, as required.

(b) Now let f ∈ L^{1}(R, μ_{Leb}). Then for any ϵ > 0 there exists a step function ϕ with ∥ϕ − f∥_{1} < ϵ. Thus we have

|f̂(t) − ϕ̂(t)| = |∫_{R}(f(x) − ϕ(x))e^{−ixt} dx|
≤ ∫_{R} |f(x) − ϕ(x)||e^{−ixt}| dx
= ∫_{R} |f(x) − ϕ(x)| dx
= ∥f − ϕ∥_{1} < ϵ,

by assumption. Now since ϕ̂(t) → 0 as t → ±∞, for any ϵ > 0 and any sequence {x_{n}} with x_{n} → ±∞, there exists N such that for all n > N we have |ϕ̂(x_{n})| < ϵ so that |f̂(x_{n})| < 2ϵ. It follows that f̂(t) → 0 as t → ±∞.

(4) Let f, g ∈ L^{1}(R, μ_{Leb}).

(a) We have

{ \atop f ∗ g}(t) = ∫_{R} ∫_{R} g(y)f(x − y)e^{−ixt} dy dx
= ∫_{R} ∫_{R} g(y)e^{−iyt} f(x − y)e^{−i(x−y)t} dy dx
= ∫_{R} ∫_{R} g(y)e^{−iyt} f(x − y)e^{−i(x−y)t} dx dy

by Fubini’s Theorem

= ∫_{R} ∫_{R} g(y)e^{−iyt} f(u)e^{−iut} du dy

by the linear change of variables x − y = u (since y is fixed in the x integral)

= f̂(t)ĝ(t),

as required.

(b) Suppose that there exists some g ∈ L^{1}(R, μ_{Leb}) satisfying f ∗ g = f for all f ∈ L^{1}(R, μ_{Leb}).

Then by part (a), we have

{ \atop f ∗ g} = f̂ĝ = f̂.

Now there exists some f ∈ L^{1}(R, μ_{Leb}) with f̂(t) ≠ 0 for all t ∈ R. Indeed, we may take

f(x) = \begin{cases}1 & if x ∈ [−π/2, π/2] \\ 0 & otherwise \end{cases} .

Then it is not difficult to check that f̂(2) = 2/t ≠ 0 for any t ∈ R. Thus, in order for the above equality to hold for all f ∈ L^{1} we must have ĝ ≡ 1, which by 3(b) is impossible as we must have ĝ(t) → 0 as t → ±∞.