$f'$ has a limit $\Longrightarrow$ $f$ has a limit without codomain Banach?

Question

Using Hahn Banach and Cauchy criterion I am able to prove the following:

Let $X$ normed, $Y$ Banach and $U \subset X$ open. Suppose that for $x_0 \in U$ we have $f : U \setminus \{x_0 \} \longrightarrow Y$ Gâteaux differentiable and such that $f'$ has a limit at $x_0$. Then $f$ has a limit at $x_0$ and whence a Gâteaux differentiable extension.

Can I relax the assumption "$Y$ Banach"?

Here is how I prove this result.

1. Using Hahn Banach extension we can show that if $[a,b]$ is a real compact interval, $Y$ normed space, $f : [a,b] \longrightarrow Y$ continuous and differentiable on $(a,b)$, then $$ ||f(b)-f(a)||_Y \leq |b-a|\sup_{(a,b)}||f'||. $$

2. If $f : U \subset X \longrightarrow Y$ is Gâteaux then the last inequality holds replacing $a,b$ by $x,y \in U$ such that $[x,y] \subset U$.

3. If $f : U \setminus \{x_0 \} \longrightarrow Y$ Gâteaux differentiable and such that $f'$ has a limit at $x_0$, to show $f$ has a limit at $x_0$ we only have to show that forall $U \setminus \{x_0 \} \ni x_j \longrightarrow x_0$, $(f(x_j))$ has a limit in $Y$. So fix $(x_j)$ sequence of $U \setminus \{x_0 \}$ with limit $x_0$, to get that $(f(x_j))$ has a limit in $Y$ it is enough to check it is Cauchy. Take $j$ large so that $x_j$ falls in a convex neighborhood of $x_0$ in $U$: $$ ||f(x_j)-f(x_{j+p})||_Y \leq ||x_j-x_{j+p}||_X \sup_{(x_j,x_{j+p})} ||f'|| $$ where the supremum is bounded because $f'$ has a limit at $x_0$. Then the $||\cdot||_X$ part is Cauchy because $(x_j)$ is convergent, so $(f(x_j))$ is Cauchy.

We only need the completeness in part 3 but I don't see how to avoid it, the limit should arise from somewhere!

Edit: the claimed fact is false because even when $X = Y = \mathbb R$ we cannot provide continuity of $f$ at $x_0$. For instance $f(x) = -1$ for $x < 0$ and $f(x) = 1$ for $x > 0$ is a counter example. This is clear from the proof: in point 3 I use the mean value inequality on $[x_j,x_{j+p}]$ but this interval may contain $x_0$ hence $f$ may fail to satisfy the Rolle's condition on this interval.

Answer 1

Here is a counterexample. Take $X=\mathbb R$ and $Y=(c_{00}, \|\cdot\|_{l^1})$, where the latter is not complete.

Let $\phi:\mathbb R\to \mathbb R$ be a continuously differentiable function such that $\phi(\mathbb R)= [0,1]$, $\phi(t)=0$ for all $t<0$, and $\phi(t)=1$ for all $t\ge 1$. Let $M>0$ such that $|\phi'(t)|\le M$ for all $t$.

Define $f:\mathbb R\setminus\{0\}\to c_{00}$ as follows: The $k$-th entry of $f(t)$ is equal to $$ f(t)_k = \frac{1}{k^5} \phi(\frac1{t^2}-k^2). $$ Then $f(t)_k=0$ for all $k \ge \frac1t$, and $f(t)\in c_{00}$. Let $t\ne0$. Then locally around $t$, the function $f$ has finite-dimensional image, and the derivative can be computed entry-wise: $$ (f'(t))_k = - \frac{2}{k^5t^3} \phi'(\frac1{t^2}-k^2). $$ Note that $\phi'(\frac1{t^2}-k^2) \ne 0$ only if $\frac1{t^2}-k^2 \in (0,1)$. Hence, we have the pointwise convergence $$ \lim_{t\to0} (f'(t))_k = 0. $$ It remains to estimate the non-zero entries of $f'(t)$. The entry $f'(t)_k$ is non-zero only if $\frac1{t^2}\le k^2+1$, which implies $t^{-3}\le (k^2+1)^{3/2}$. Hence, the non-zero entries of $f'(t)$ can be estimated by $$ |f'(t)_k| \le \frac{2 M}{k^5t^3} \le \frac{2 M(k^2+1)^{3/2}}{k^5}, $$ which is a summable dominating sequence. Hence by dominated convergence $f'(t)\to 0$ in $l^1$.

We have the entry-wise limit $\lim_{t\to 0} f(t)_k=\frac1{k^5}$, which is not an element of $c_{00}$. And $\lim_{t\to0} f(t) $ does not exist.

Answer 2

Here is a counterexample. Let $c_0$ be the Banach space of null sequences (with supremum norm) and $c_c$ the (incomplete) subspace of finite sequences.

Let $\varphi \colon \mathbb R \to \mathbb R$ be smooth with $\varphi(t) = 0$ for $t \le 0$ and $\varphi(t) = 1$ for $t \ge 1$.

Now, we define $f \colon \mathbb R \to c_0$ via $$ f(t)_n = \frac1{n^3} \varphi( n (n-1)(1/n-|t|)). $$ Note that $f(t)_n = 0$ for $|t| \ge 1/(n-1)$ and $f(t)_n = 1/n^3$ for $|t|\le1/n$. Then, we can check

• $f(t) \in c_c$ for $t \ne 0$ and $f(0) \in c_0 \setminus c_c$
• $f$ is everywhere Gâteaux differentiable
• $f'$ is continuous with $f'(0) = 0$
• $f'(t) \in c_c$ for all $t \in \mathbb R$

Thus, we get a counterexample if we consider $f$ as a function from $\mathbb R \setminus \{0\}$ to $Y = c_c$.