Let f ∈ C ∞ ( [ 0 , ∞ ) × [ 0 , 1 ] ) f \in C^\infty\p{\pco{0, \infty} \times \br{0, 1}} f ∈ C ∞ ( [ 0 , ∞ ) × [ 0 , 1 ] )
∫ 0 ∞ ∫ 0 1 ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d x d t < ∞ . \int_0^\infty \int_0^1 \abs{\partial_t f\p{t, x}}^2 \p{1 + t^2} \,\diff{x} \,\diff{t} < \infty. ∫ 0 ∞ ∫ 0 1 ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d x d t < ∞. Prove that there exists a function g ∈ L 2 ( [ 0 , 1 ] ) g \in L^2\p{\br{0,1}} g ∈ L 2 ( [ 0 , 1 ] ) f ( t , ⋅ ) f\p{t, \:\cdot\:} f ( t , ⋅ ) g ( ⋅ ) g\p{\:\cdot\:} g ( ⋅ ) L 2 ( [ 0 , 1 ] ) L^2\p{\br{0,1}} L 2 ( [ 0 , 1 ] ) t → ∞ t \to \infty t → ∞
Solution.
Because L 2 ( [ 0 , 1 ] ) L^2\p{\br{0,1}} L 2 ( [ 0 , 1 ] ) { x ↦ f ( t , x ) } t \set{x \mapsto f\p{t, x}}_t { x ↦ f ( t , x ) } t L 2 ( [ 0 , 1 ] ) L^2\p{\br{0,1}} L 2 ( [ 0 , 1 ] )
Let t 1 ≤ t 2 t_1 \leq t_2 t 1 ≤ t 2
∫ t 1 t 2 ∣ ∂ t f ( t , x ) ∣ d t = ∫ 0 ∞ χ [ t 1 , t 2 ] 1 + t 2 ∣ ∂ t f ( t , x ) ∣ 1 + t 2 d t = ( ∫ t 1 t 2 1 1 + t 2 d t ) 1 / 2 ( ∫ 0 ∞ ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d t ) 1 / 2 ≤ ( arctan t 2 − arctan t 1 ) 1 / 2 ( ∫ 0 ∞ ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d t ) 1 / 2 . \begin{aligned}
\int_{t_1}^{t_2} \abs{\partial_t f\p{t, x}} \,\diff{t}
&= \int_0^\infty \frac{\chi_{\br{t_1,t_2}}}{\sqrt{1 + t^2}} \abs{\partial_t f\p{t, x}}\sqrt{1 + t^2} \,\diff{t} \\
&= \p{\int_{t_1}^{t_2} \frac{1}{1 + t^2} \,\diff{t}}^{1/2} \p{\int_0^\infty \abs{\partial_t f\p{t, x}}^2\p{1 + t^2} \,\diff{t}}^{1/2} \\
&\leq \p{\arctan{t_2} - \arctan{t_1}}^{1/2} \p{\int_0^\infty \abs{\partial_t f\p{t, x}}^2\p{1 + t^2} \,\diff{t}}^{1/2}.
\end{aligned} ∫ t 1 t 2 ∣ ∂ t f ( t , x ) ∣ d t = ∫ 0 ∞ 1 + t 2 χ [ t 1 , t 2 ] ∣ ∂ t f ( t , x ) ∣ 1 + t 2 d t = ( ∫ t 1 t 2 1 + t 2 1 d t ) 1/2 ( ∫ 0 ∞ ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d t ) 1/2 ≤ ( arctan t 2 − arctan t 1 ) 1/2 ( ∫ 0 ∞ ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d t ) 1/2 . Then by the fundamental theorem of calculus,
∥ f ( t 1 , ⋅ ) − f ( t 2 , ⋅ ) ∥ L 2 2 = ∫ 0 1 ∣ f ( t 1 , x ) − f ( t 2 , x ) ∣ 2 d x ≤ ∫ 0 1 ( ∫ t 1 t 2 ∣ ∂ t f ( t , x ) ∣ d t ) 2 d x ≤ ( arctan t 2 − arctan t 1 ) ∫ 0 1 ∫ 0 ∞ ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d t d x = ( arctan t 2 − arctan t 1 ) ∫ 0 ∞ ∫ 0 1 ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d x d t . ( Fubini-Tonelli ) \begin{aligned}
\norm{f\p{t_1, \:\cdot\:} - f\p{t_2, \:\cdot\:}}_{L^2}^2
&= \int_0^1 \abs{f\p{t_1, x} - f\p{t_2, x}}^2 \,\diff{x} \\
&\leq \int_0^1 \p{\int_{t_1}^{t_2} \abs{\partial_t f\p{t, x}} \,\diff{t}}^2 \,\diff{x} \\
&\leq \p{\arctan{t_2} - \arctan{t_1}} \int_0^1 \int_0^\infty \abs{\partial_t f\p{t, x}}^2\p{1 + t^2} \,\diff{t} \,\diff{x} \\
&= \p{\arctan{t_2} - \arctan{t_1}} \int_0^\infty \int_0^1 \abs{\partial_t f\p{t, x}}^2\p{1 + t^2} \,\diff{x} \,\diff{t}.
&& (\text{Fubini-Tonelli})
\end{aligned} ∥ f ( t 1 , ⋅ ) − f ( t 2 , ⋅ ) ∥ L 2 2 = ∫ 0 1 ∣ f ( t 1 , x ) − f ( t 2 , x ) ∣ 2 d x ≤ ∫ 0 1 ( ∫ t 1 t 2 ∣ ∂ t f ( t , x ) ∣ d t ) 2 d x ≤ ( arctan t 2 − arctan t 1 ) ∫ 0 1 ∫ 0 ∞ ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d t d x = ( arctan t 2 − arctan t 1 ) ∫ 0 ∞ ∫ 0 1 ∣ ∂ t f ( t , x ) ∣ 2 ( 1 + t 2 ) d x d t . ( Fubini-Tonelli ) The integral is bounded by assumption, so this tends to 0 0 0 t 1 , t 2 → ∞ t_1, t_2 \to \infty t 1 , t 2 → ∞ arctan t → π 2 \arctan{t} \to \frac{\pi}{2} arctan t → 2 π { x ↦ f ( t 1 , x ) } t \set{x \mapsto f\p{t_1, x}}_t { x ↦ f ( t 1 , x ) } t L 2 ( [ 0 , 1 ] ) L^2\p{\br{0,1}} L 2 ( [ 0 , 1 ] )