Let $A:X\to Y$, where $X=C^1[0,1]$ equipped with the norm $\|f\|_{C^1}=\|f\|_u+\|f'\|_u$, and $Y=C[0,1]$ equipped with the uniform norm $\|f\|_u$ be defined as $$(Af)(x)=f'(x);$$ I proved the linearity as follows: $A(f+g)(x)=(f+g)'(x)=f'(x)+g'(x)=Af(x)+Ag(x);$ $A(\lambda f)(x)=(\lambda f)'(x)=\lambda f'(x)=\lambda A f(x)$ I do not understand how to prove continuity and find the norm
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1On what spaces is your operator acting? With what norms? Please post all relevant details of the problem. – Jose27 Apr 10 '22 at 17:00
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This is not a good way to ask a question. First you need to specify the domain and codomain of the map $A$. – Sourav Ghosh Apr 10 '22 at 17:01
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@SuccessfulFailure Corrected, wrote everything that was in the condition – Andrij Matviiv Apr 10 '22 at 17:03
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@Jose27 Corrected – Andrij Matviiv Apr 10 '22 at 17:03
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@AndrijMatviiv try to observe the continuity at $0$.Can you produce a sequence of continuously differentiable function $(f_n)$ that converges to $0$ uniformly but $A(f_n) $ doesn't converges to $0$ ? – Sourav Ghosh Apr 10 '22 at 17:06
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@SuccessfulFailure I don't understand you a little – Andrij Matviiv Apr 10 '22 at 17:09
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1You've gotten two conflicting answers because you've not defined the norm on $X$. – Teepeemm Apr 11 '22 at 02:00
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Hint :
$T\in \mathcal{L}(X, Y) $ is continuous iff $T$ is continuous at $0_X$. (i.e a linear map between two normed spaces either continuous everywhere or discontinuous everywhere.)
$T\in \mathcal{L}(X, Y) $ is continuous at $0$ iff for every sequence $(x_n) \subset X$ with $x_n\to 0 \implies Tx_n \to 0 $
Now, for your question choose,
$f_n(t) =\frac{1}{n} \sin (nt) $
Then try to show :
$(f_n) \subset C^1[0, 1] $
$f_n \to 0 $ uniformly on $[0, 1]$
$A(f_n) $ doesn't converge to $ 0 $ uniformly on $[0, 1]$.
Remember that operator norm is defined only when the operator is continuous (i.e bounded).
Sourav Ghosh
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1Yes, I understand, but if we talk about the norm, how to find it here? It's just hard to imagine here, although maybe one – Andrij Matviiv Apr 10 '22 at 17:27
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1Remember that operator norm is defined only when the operator is continuous (i.e bounded). – Sourav Ghosh Apr 10 '22 at 17:36
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1$|T|=\sup{|Tx|:|x|\le 1}$ \For unbounded operator $T$ , the set ${|Tx|:|x|\le 1}$ is not bounded above and $\sup$ doesn't exist. – Sourav Ghosh Apr 10 '22 at 17:37
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The key is the norms on $X,Y$. If $\|f\|_\infty = \sup_{x \in [0,1]}|f(x)|$, then $\|f\|_X = \|f\|_\infty+ \|f'\|_\infty$ and $\|f\|_Y = \|f\|_\infty$.
Hence $\|Af\|_Y = \|f'\|_Y \le \|f\|_X$, and so $A$ is continuous and the norm bounded by one.
Consider $f_n(x) = x^n$ in order to show that this bound is tight.
copper.hat
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Why the downvote? If there is something wrong I would like to correct it. – copper.hat May 22 '22 at 00:45