Continuity for a nonlinear functional

continuityfunctional-analysisnonlinear-analysis

I'd like to check the continuity for the nonlinear functional $T: (C^{0}([0,1],\Vert \cdot \Vert_{\infty}) \rightarrow (C^{0}([0,1],\Vert \cdot \Vert_{\infty}) $ , with $T(f)(x)=\arctan(f(x))$.

I need to show that for $f_n \rightarrow f$ in infinity norm, then $T(f_n) \rightarrow T(f)$, also in infinity norm.

So, I need to compute $\sup_{x \in [0,1]} | \arctan(f(x))- \arctan(f_k(x))|$, and I want to show it goes to zero, but how can I compute it formally?

Best Answer

$|\operatorname{arctan} (a) -\operatorname{arctan}(b)| \leq |a-b|$ by MVT so $Tf_n \to Tf$ uniformly.

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Sequence of continuously differentiable functions

The argument in d) should be the following way.

We have \begin{align*} \sup_{x\in[0,1]}|I(f)'(x)|=\sup_{x\in[0,1]}\left|\dfrac{d}{dx}\int_{0}^{x}f(t)dt\right|=\sup_{x\in[0,1]}|f(x)|, \end{align*} and hence \begin{align*} \|f\|_{1}\leq\sup_{x\in[0,1]}\int_{0}^{x}|f(t)|dt+\sup_{x\in[0,1]}|I(f)'(x)|\leq 2\|f\|_{\infty}, \end{align*} so $I$ is continuous.

Continuity of $T(f) = \int (h\circ f)\operatorname df$

The answer is that if we consider the particularization of the $n$-dimensional BV norm for $n=1$, i.e. $$\DeclareMathOperator{\Dm}{d\!} \Vert f\Vert_{BV} \triangleq \Vert f\Vert_{L_1}+\operatorname{TV}(f)\label{1}\tag{1} $$ where $\operatorname{TV}(f)$ is the Tonelli-Cesari total variation as given in the OP and in my comment, the operator $T(f)$ is discontinuous.
On the other hand, if we consider the classical $1$-dimensional BV norm $$ \Vert f\Vert_\mathscr{BV} \triangleq \Vert f\Vert_{L_1}+\operatorname{\mathscr{V}}(f)\label{2}\tag{2} $$ where $$ \operatorname{\mathscr{V}}(f)=\sup_{P \in \mathscr{P}} \sum_{i=0}^{n_{P}-1} | f(x_{i+1})-f(x_i) | $$ is the classical (or we may say Jordan) variation of $f$ and $\mathscr{P}$ is the set of finite partitions of $\Bbb R$, then $T(f)$ is continuous.

Proof of the discontinuity of $T(f)$ respect to $\Vert \cdot\Vert_{BV}$. Consider $f$ as the characteristic function of the open interval $]0, 1[$, i.e. $$ f(x) \triangleq\chi_{]0,1[}(x)= \begin{cases} 1 & x\in ]0,1[,\\ 0 & x\notin ]0,1[. \end{cases} $$ As we can easily prove, $f\in BV(\Bbb R, [0,1])$: now let's define a sequence $\{f_n\}_{n\in\Bbb N_{>0}}\subsetneq BV(\Bbb R, [0,1])$ converging to $f$ as $$ f_n(x)\triangleq \begin{cases} \varphi_n\ast f(x) = \displaystyle\int\limits_{\Bbb R}\varphi_n(x-y)f(y)\Dm y = \int\limits_0^1\varphi_n(x-y)\Dm y &x\neq 0,\\ 1-\frac{1}{2}\sin\frac{1}{n} & x=0. \end{cases} $$ where $\varphi_n(x)$ is any sequence of $C^\infty_0$ mollifiers converging to the Dirac measure as $n\to\infty$: note that also $f_n(x)\to f(x)$ pointwise for all $x\in\Bbb R\setminus\{0\}$, while for $x=0$ the pointwise limit does not exist due to the rapid oscillations of the function sequence at that point.
Now it's easy to see that $$ \Dm f = \delta(0)-\delta(1) = \lim_{n\to\infty} \Dm f_n\;\text{ in the space of Radon measures }\;\mathcal{M}(\Bbb R).\label{3}\tag{3} $$ and while this implies that $$ \int (h\circ f) \Dm f = h\circ f(0) - h\circ f(1) =h(0)-h(0)=0 $$ it also implies that $T(f_n)\nrightarrow 0$ as $n\to\infty$ since for a sufficiently large $n$ we have that $$ \begin{split} T(f_n) &= \int (h\circ f_n) \Dm f_n = \int\limits_{\Bbb R} (h\circ f_n) \Dm f_n \\ & = \int\limits_{\Bbb R} (h\circ f_n) \Dm f_n + h \circ f_n(0) - h \circ f_n(1) - h \circ f_n(0) + h \circ f_n(1)\\ & = h \circ f_n(0) - h \circ f_n(1) \\ & \qquad\qquad+ \int\limits_{-\infty}^{+\frac 12} \big(h\circ f_n- h \circ f_n|_{x=0}\big) \Dm f_n + \int\limits_{1\over 2}^{+\infty} \big(h \circ f_n - h\circ f_n|_{x=1}\big) \Dm f_n\\ & = h \left(1-\frac{1}{2}\sin\frac{1}{n} \right) - h \circ f_n(1)\\ & \qquad\qquad + \int\limits_{-\infty}^{+\frac 12} \big(h\circ f_n- h \circ f_n|_{x=0}\big) \Dm f_n + \int\limits_{1\over 2}^{+\infty} \big(h \circ f_n - h\circ f_n|_{x=1}\big) \Dm f_n \end{split}\label{4}\tag{4} $$ Now while the last three terms on the right side converge to a limit for $n\to\infty$, precisely $$ \begin{align} \lim_{n\to\infty} & h \circ f_n(1) = h(0), \\ \lim_{n\to\infty} & \int\limits_{-\infty}^{+\frac 12} \big(h\circ f_n- h \circ f_n|_{x=0}\big) \Dm f_n = 0, \\ \lim_{n\to\infty} & \int\limits_{1\over 2}^{+\infty} \big(h \circ f_n - h\circ f_n|_{x=1} \big) \Dm f_n = 0 \end{align} $$ the first term, i.e. $h \left(1-\frac{1}{2}\sin\frac{1}{n} \right)$ does not, thus $\vert T(f_n) - T(f)\vert \nrightarrow 0$ for $n\to\infty$ and the functional $T:{BV}(\Bbb R, [0,1]) \to \Bbb R$ is not continuous respect to the topology of ${BV}(\Bbb R, [0,1])$.

Proof of the continuity of $T(f)$ respect to $\Vert \cdot \Vert_\mathscr{BV}$. This is simply a consequence of the fact that $\Vert \cdot \Vert_\mathscr{BV}$ forces any converging sequence in the corresponding Banach space $\mathscr{BV}(\Bbb R, [0,1])$ to converge also pointwisely.

Final notes.

In the development above, for the sake of clarity I've adopted some non standard notations: precisely in the literature $\mathscr{V}(f)$ is $V(f)$ and we speak of BV spaces with norm $\|\cdot\|_{BV}$ defined by \eqref{1} or \eqref{2}, leaving to the analysis of the context the understanding of what of the two is effectively used.
The problem (as in this Q&A) is exactly the one also pointed out by Liding Yao in his comment: $ \{f_n\}_{n\in\Bbb N_{>0}}$ should be pointwise known on the support of the singular part of the Radon measure defined by its limit $f$, and this cannot be guaranteed if you are working in BV.
Perhaps a clearer explanation of the passages of why relations \eqref{3} and \eqref{4} hold is needed. These equations are consequences of the fact that $\{\varphi_n\}_{n\in}\subsetneq C_o^\infty(\Bbb R)$ is a sequence of smooth mollifiers.

Best Answer

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Sequence of continuously differentiable functions

Continuity of $T(f) = \int (h\circ f)\operatorname df$

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