# Showing map is Frechet differentiable (will add bounty)

Define the seminorm $$[u]_{\alpha} = \sup_{(x,t), (y,s) \in Q} \frac{|u(x,t) - u(y,s)|}{(|x-y|^2 + |t-s|)^{\frac{\alpha}{2}}},$$ and the norm $$\lVert u\rVert_{{C}^{k, \alpha}(\overline{Q})} = \sum_{i+2j \leq k} \left\lVert \frac{\partial^{i+j}u}{\partial x^i \partial t^j}\right\rVert_{C(\overline{Q})} + \sum_{i+2j = k} \bigg[\frac{\partial^{i+j}u}{\partial x^i \partial t^j}\bigg]_\alpha$$ composed of the seminorm above and the usual sup norm on continuous functions.

I want to show that the map $f:C^{k+2, \alpha} \to C^{k, \alpha}$ defined by $$f(u) = u_t - g(x,t,u,u_x),$$ where $g$ is smooth in all its arguments, is Frechet differentiable at the point $v = g(x,t,0,0)t$. I know that the directional derivative of $f$ at $v$ is $$df(v)h = h_t - (g_z|_{v}h + g_p|_{v}h_x)$$ where $g(x,t,z,p)$ are the arguments.

So I need to prove that $$\lim_{h \to 0}\frac{\lVert f(v+h)-f(v) - h_t - (g_z|_{v}h + g_p|_{v}h_x) \rVert_{C^{k,\alpha}}}{\lVert h\rVert_{C^{k+2,\alpha}}} = 0.$$ The LHS is $$\lim_{h \to 0}\frac{ \lVert g(x,t,v, v_x) - g(x,t, v+ h, v_x + h_x) + g_z|_v h + g_p|_vh_x \rVert_{C^{k,\alpha}}}{\lVert h\rVert_{C^{k+2,\alpha}}}$$

which is a horrible thing to simplify. How can I prove that the limit is 0? I need serious help please. I will add a bounty when to a good answer. Thanks.

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One approach is to show that $df(v)$ is continuous with respect to $v$, that is, $\|df(v)-df(v_1)\|$ is small when $\|v-v_1\|$ is small. The continuity of $df$ implies that it is actually the Frechet derivative, see here. –  user31373 Jul 11 '12 at 0:08
@LeonidKovalev That also seems hard to prove :| –  Court Jul 12 '12 at 16:02
There are three parts of $df(v)=S+T+U$. One is $Sh=h_t$, this does not depend on $v$ at all. The second is $Th=g_z(v)h$, where $g_z$ is continuous in $v$. The third is $Uh=g_p(v)h_x$, where $g_p$ is also continuous in $v$. –  user31373 Jul 12 '12 at 16:13
@LeonidKovalev Thanks. So if $v_n \to v$ in the $C^{k+2, \alpha}$ norm then $df(v_n)h \to df(v)h$ because of what you wrote above. Hence $df(v_n) \to df(v)$ which finishes it. I'm happy to give you the bounty if you write your comment as an answer. –  Court Jul 12 '12 at 18:10
Be careful with convergence of linear functionals: you want the norm convergence $\|df(v_n)-df(v)\|\to 0$, not just pointwise (weak) convergence. –  user31373 Jul 12 '12 at 18:37