I am stuck trying to reach an (in)equality...
Let $\Omega \in \mathbb{R}$ and $f=f(t,x): \mathbb{R} \supset[0,T] \times \Omega \rightarrow \mathbb{R}$ be an element of $$\mathcal{W}:=L_{_t}^2(0,T,H_{_x}^1(\Omega)):= \left\{ \varphi:[0,T]\times \Omega \rightarrow \mathbb{R} \, \Big| \, \int_{_{[0,T]}} \| \varphi (t,x) \|_{_{H_{_x}^1(\Omega)}}^{^2} dt < \infty \right\}.$$
I want to reach something like

$$ \frac{d}{dt}\Big\langle f(t,x),f(t,x) \Big\rangle_{_{\mathcal{W}}} \leq \Big\langle \dot f(t,x),f(t,x)\Big\rangle_{_{\mathcal{W}}}$$

to then extend and obtain

$$ \frac{d}{dt} \Big\|f(t,x) \Big\|_{_{\mathcal{W}}} = \frac{\tfrac{d}{dt}\Big\|f(t,x) \Big\|_{_{\mathcal{W}}}\Big\|f(t,x) \Big\|_{_{\mathcal{W}}}}{\Big\|f(t,x) \Big\|_{_{\mathcal{W}}}}\leq \frac{\Big\langle \dot f(t,x),f(t,x)\Big\rangle_{_{\mathcal{W}}}}{\Big\|f(t,x) \Big\|_{_{\mathcal{W}}}} $$

is this just straight forward? what am I missing?

  • 2
    $\begingroup$ Please explain your notation. What does the inner product mean? What is a vector here? If you really just mean $\int_0^T f(t)^2dt$, then this is a constant and its derivative with respect to $t$ is 0, as it does not depend on $t$. The "$tdt$" are dummy variables that are integrated out. $\endgroup$ – Michael Jun 2 '16 at 18:50
  • $\begingroup$ ok. I'll clarify. I missed to see that. ... did you voted the question down? $\endgroup$ – scjorge Jun 2 '16 at 18:53
  • $\begingroup$ No I did not downvote. $\endgroup$ – Michael Jun 2 '16 at 18:54
  • $\begingroup$ ok. it seems rude for me if someone votes down without commenting or something. anyway... I extended the question. my function is $f(t,x)$ or $f(t,\cdot): \Omega \rightarrow \mathbb{R}$ $\endgroup$ – scjorge Jun 2 '16 at 18:58
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    $\begingroup$ In that case, assuming a real-inner project space, you can make progress via $$\langle f(t+h),f(t+h) \rangle - \langle f(t), f(t)\rangle = \langle f(t+h)-f(t), f(t+h)\rangle + \langle f(t), f(t+h)\rangle - \langle f(t) , f(t) \rangle$$ $\endgroup$ – Michael Jun 2 '16 at 19:13

For a real Hilbert space $H$, the square norm $N(h):=\langle h , h ⟩$ is (Fréchet) differentiable with derivative $dN(h)v = 2⟨ h, v ⟩ $. In this case we have $H=L^2_x$ where the subscript indicates that its the space of spatially $L^2$ functions. We then have the chain rule for functions $f=f(t,x)∈ H^1_tL^2_x = H^1([0,T],L^2(Ω)) = H^1(0,T;L^2(Ω))$, $$ \frac{d}{dt}‖f(t)‖_{L^2_x}^2 = \frac{d}{dt} (N \circ f)(t) = dN(f(t))f'(t) = 2⟨ f(t),f'(t)⟩ $$ At the same time, we have $\frac{d}{dt}‖f(t)‖_{L^2}^2 = 2‖f(t)‖_{L^2} \frac{d}{dt}‖f(t)‖_{L^2}$ by chain rule and the identity $(x^2)'= 2x$. Hence we have that $$\frac{d}{dt} ‖f(t)‖_{L^2} = \left\langle \frac{f(t)}{‖f(t)‖} , f'(t)\right\rangle$$

Note well that this is for functions in a different space than the one you prescribed $L^2_tH^1_x$,

$$L^2([0,T],H^1(Ω)) = \left\{ f:[0,T]\times Ω → \Bbb R \ \middle|\ \substack{ \displaystyle f(t),\partial_xf(t) ∈ L^2(dx) \text{ for a.e. }t\\ \displaystyle ∫_0^T ‖f‖^2_{L^2} + ‖\partial_x f‖^2_{L^2} \ dt< ∞} \right\}$$

The space we use is instead $H^1_xL^2_t = H^1([0,T],L^2(Ω))$, which is defined as $$H^1([0,T],L^2(Ω)) = \left\{ f:[0,T]\times Ω → \Bbb R \ \middle|\ \substack{ \displaystyle f(t),\partial_tf(t) ∈ L^2(dx) \text{ for a.e. $t$}\\ \displaystyle ∫_0^T ‖f‖^2_{L^2} + ‖\partial_t f‖^2_{L^2} \ dt< ∞} \right\}$$

(strictly speaking we don't need the top line in the definition since its implied by writing the second line.)

  • $\begingroup$ this seems to be what I need. thanx. could you please also write explicitly the definition of the space $H^1(0,T;L^2(\Omega))$? for comparison. the $L^2_x$-Notation is still new to me... $\endgroup$ – scjorge Jun 2 '16 at 20:03
  • $\begingroup$ @scjorge done :) $\endgroup$ – Calvin Khor Jun 2 '16 at 20:20
  • $\begingroup$ Thanx you! just one last question, what is $L^2(dx)$? $\endgroup$ – scjorge Jun 2 '16 at 20:51
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    $\begingroup$ @scjorge its the linear operator $dN(h) = (dN)(h)(· ) = ⟨ h, · ⟩ $ applied to $v$. $\endgroup$ – Calvin Khor Jun 3 '16 at 14:17
  • 1
    $\begingroup$ @scjorge ah damnit yes :P $\endgroup$ – Calvin Khor Jun 3 '16 at 14:33

I've found two ways to reach an inequality similar to the one asked.

1. Similar to @CalvinKhor's answer but more precise:

For a real Hilbert space $H$, the norm $ \big\| \,\cdot\, \big\|_{_{H}} $ is Fréchet-differentiable.
Its derivative at, say, $0 \neq v \in H$ is $$\big(D\big\| \,\cdot\, \big\|_{_{H}} \big)\,v = \frac{⟨ \,\cdot\, , v\, ⟩_{_{H}}}{\| \,\cdot\, \|_{_{H}}}. $$ This is from Dirk Werner's Funktionalanalysis. Chapter III, section 5.
He uses the chain rule and the "identity" $$ D(\,\cdot\,)^2 = 2 \times \cdot \,$$ (the derivative of the "square-function" is the "double-function")
With this, we get for $\mathcal{W}=L^{^2}_{_t}(0,T,H^{^1}_{_{x}}(\Omega))$ and $0\neq v = \dot f(t,x)$ $$ \Big(D\big\| f(t,x) \big\|_{_{\mathcal{W}}} \Big)\,\big(\dot f(t,x)\big) = \frac{\Big\langle f(t,x) , \dot f(t,x)\, \Big\rangle_{_{\mathcal{W}}}}{\| f(t,x) \|_{_{\mathcal{W}}}}= \frac{\Big\langle \dot f(t,x)\, , f(t,x) \, \Big\rangle_{_{\mathcal{W}}}}{\| f(t,x) \|_{_{\mathcal{W}}}}. $$

2. Use the linear approximation of a real-valued differentiable function, the reverse triangle inequality, and an inner-product-space-structure.

We may write the differential quotient of the norm of a time function as: \begin{align*} \frac{d}{dt} \big\| f(t) \big\| &:= \lim_{h \rightarrow 0} \frac{\big\|f(t+h)\big\|-\big\|f(t)\big\|}{h} \\ &= \lim_{h \rightarrow 0} \frac{\big\|f(t+h)\big\|-\big\|f(t)\big\|}{h}\cdot\frac{\big\|f(t+h)\big\|+\big\|f(t)\big\|}{\big\|f(t+h)\big\|+\big\|f(t)\big\|} \\ &= \lim_{h \rightarrow 0} \frac{\big\|f(t+h)\big\|^2-\big\|f(t)\big\|^2}{h\big(\big\|f(t+h)\big\|+\big\|f(t)\big\|\big)} \\ &= \lim_{h \rightarrow 0} \frac{\big\|f(t) + f'(t)\cdot h + o(|h|) \big\|^2-\big\|f(t)\big\|^2}{h\big(\big\|f(t)+f'(t)\cdot h + o(|h|)\big\|+\big\|f(t)\big\|\big)}. \end{align*} If we then apply the inverse triangle inequality: $$ \|x\| - \|y\| \leq \Big| \|x\| - \|y\| \Big| \leq \big\| x \pm y \big\| $$ one gets: $$ \frac{1}{\big\|f(t) + f'(t)\cdot h + o(|h|)\big\|} \leq \frac{1}{\big\|f(t)\big\| - \big\|f'(t)\cdot h + o(|h|)\big\|}. $$ Thus we have: $$ \frac{d}{dt} \big\| f(t) \big\| \leq \lim_{h \rightarrow 0} \frac{\big\|f(t) + f'(t)\cdot h + o(|h|) \big\|^2-\big\|f(t)\big\|^2}{h\big(\big\|f(t)\big\|-\big\|f'(t)\cdot h + o(|h|)\big\|+\big\|f(t)\big\|\big)} \\ $$ Given that the norm is induced by an inner product, we expand: \begin{align*} \frac{d}{dt} \big\| f(t) \big\| &\leq \lim_{h \rightarrow 0} \frac{\big\langle f(t) + f'(t)\cdot h + o(|h|),\,f(t) + f'(t)\cdot h + o(|h|) \big\rangle-\big\|f(t)\big\|^2}{h\big(\big\|f(t)\big\|-\big\|f'(t)\cdot h + o(|h|)\big\|+\big\|f(t)\big\|\big)} \\ &\leq \lim_{h \rightarrow 0} \frac{\big\| f(t) \big\|^2 + 2 \big\langle f(t),\,f'(t)\cdot h + o(|h|) \big\rangle + \big\| f'(t)\cdot h + o(|h|) \big\|^2 -\big\|f(t)\big\|^2}{h\big(2 \big\|f(t)\big\|-\big\|f'(t)\cdot h + o(|h|)\big\|\big)} \\ &\leq \lim_{h \rightarrow 0} \frac{ 2 \big\langle f(t),\,f'(t)\cdot h + o(|h|) \big\rangle + \big\| f'(t)\cdot h + o(|h|) \big\|^2 }{h\big(2\big\|f(t)\big\|-\big\|f'(t)\cdot h + o(|h|)\big\|\big)} \\ &\leq \lim_{h \rightarrow 0} \frac{h}{h}\cdot\frac{ 2 \big\langle f(t),\,f'(t) + \tfrac{o(|h|)}{h} \big\rangle + |h|\big\| f'(t) + \tfrac{o(|h|)}{h} \big\|^2 }{2\big\|f(t)\big\|-|h|\big\|f'(t) + \tfrac{o(|h|)}{h}\big\|}. \end{align*} Finally, the value of the limit is: \begin{align*} \frac{d}{dt} \big\| f(t) \big\| &\leq 1 \cdot\frac{ 2 \big\langle f(t),\,f'(t) + 0 \big\rangle + 0 \big\| f'(t) + 0 \big\|^2 }{2\big\|f(t)\big\|-0\big\|f'(t) + 0\big\|} \\ &\leq \frac{ 2 \big\langle f(t),\,f'(t)\big\rangle }{2\big\|f(t)\big\|} = \frac{\big\langle f(t),\,f'(t)\big\rangle}{\big\|f(t)\big\|}. \end{align*} Thus: $$ \frac{d}{dt} \big\| f(t,x) \big\|_{_{\mathcal{W}}} \leq \frac{\big\langle f(t,x),\,\dot f(t,x)\big\rangle_{_{\mathcal{W}}}}{\big\|f(t,x)\big\|_{_{\mathcal{W}}}}= \frac{\big\langle \dot f(t,x),\,f(t,x)\big\rangle_{_{\mathcal{W}}}}{\big\|f(t,x)\big\|_{_{\mathcal{W}}}}. $$

Depending on your needs, use the first or the second (-;


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