Let $C[0,1]$ be the normed linear space with norm $\Vert f \Vert_\infty=\displaystyle\sup_{x\in[0,1]}\vert f(x)\vert$ and $Y$ be the vector subspace defined by $Y:=\Big\{f\in C[0,1]:f$ is differentiable, $f'$ is continuous$\Big\}$ with norm $\Vert f\Vert_1=\displaystyle\sup_{x\in[0,1]}\vert f(x)\vert + \sup_{x\in[0,1]}\vert f'(x)\vert$. Which of the following statements are true?
$a)$ $(Y,\Vert .\Vert_\infty)$ is Banach.
$b)$ $(Y,\Vert .\Vert_1)$ is Banach
$c)$ The map $T:(Y,\Vert .\Vert_1) \to (C[0,1],\Vert .\Vert_\infty)$ by $T(f):=f'$ is continuous.
$d)$ The map $I:(C[0,1],\Vert .\Vert_\infty) \to (Y,\Vert .\Vert_1)$ by $\displaystyle I(f):=\int_0^x f(t)dt$ is continuous.
$Y$ being the $C^1[0,1]$ space with $C^1$ norm is complete hence $b)$ is TRUE.
In general, sequence of continuously differentiable functions need not converge uniformly to a continuously differentiable function. Counter-example:
$\displaystyle f_n(x)=\frac{\sin nx}{n}\xrightarrow[]{Uniformly}f\equiv0$ on $[0,1]$ but its derivative $f_n'(x)=\cos nx$ does not converge. Since the limit of derivative is not equal to derivative of the limit, thus derivative is not continuous. $a)$ is FALSE.
For $c)$, let $\{f_n\}\to f$ in $Y$ i.e. $\Vert f_n-f\Vert_1<\varepsilon$ for sufficient $n$, $$\vert (T(f_n)-T(f))(x)\vert\leq\Vert T(f_n)-T(f) \Vert_\infty=\Vert f'_n-f' \Vert_\infty\leq\Vert f_n-f \Vert_1<\varepsilon.$$
Hence $T$ is continuous. $c)$ is TRUE.
For $d)$, let $\{f_n\}\to f$ in $C[0,1]$ i.e. $\Vert f_n-f\Vert_\infty<\varepsilon$ for sufficient $n$,
$$\vert (I(f_n)-I(f))(x)\vert\leq\displaystyle \Bigg\vert \int_0^x f_n(t)-f(t)\; dt \Bigg\vert \leq\int_0^x \Vert f_n-f\Vert_\infty \; |dt|<\varepsilon.$$
Hence $T$ is continuous. $d)$ is TRUE.
Can anyone validate my arguments?
What are other counter-examples for $a)$ being false?
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Update: Infact I realised sequence of continuously differentiable function need not converge to a continuous limit.
$f_n(x)=x^n\to f=\begin{cases}0 &x\in[0,1)\\1 &x=1\end{cases}$ for $x\in[0,1]$. So $a)$ need not be TRUE.
Best Answer
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.