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I came across this question in my self study of a little fuctional analysis and I'd help in solving it. I have also attempted it in the hope that I'll be corrected where I made mistakes. Thanks.

Let $X$ and $Y$ be normed linear spaces and $T:X\to Y$ be linear. I want to show the following:

(a) $T$ is continuous if and only if it is continuous at a single point $x_0$ in $X$.
(b) $T$ is Lipschitz if and only if it is continuous.
(c) Neither (i) nor (ii) hold in the absence of the linearity assumption on $T$.

Attempts.

(a)
$(\Rightarrow)$ Obvious, since if $T$ is continuous everywhere in $X$, then it is certainly continuous at a point $x_0\in X$.

$(\Leftarrow)$ Suppose that $T$ is continuous at $x_0 \in X$. Let $\varepsilon \gt 0$ be given. Then there is a $\delta\gt 0$ such that $\|T(x)-T(x_0)\|\lt \varepsilon$ whenever $\|x-x_0\|\lt \delta.$ Let $x_1$ be any arbitrary point in $X$. Then for $\|x-x_1\| = \|(x + x_0 -x_1)-x_0\|\lt \delta$, \begin{align*} \|T(x)-T(x_1)\| & = \|T(x)+T(x_0)-T(x_1)-T(x_0)\| \\ & = \|T(x + x_0 -x_1)-T(x_0)\|\qquad(\because T ~\text{is linear})\\ & \lt \varepsilon. \end{align*} (b)
$(\Rightarrow)$ Suppose $T$ is continuous. Then $T$ is bounded. So there exists an $M\geqslant 0$ such that for $x,y\in X$ $\|T(x)-T(y)\| \leq M \|x-y\|,$ which is precisely the condition for $T$ to be Lipschitz.

$(\Leftarrow)$ Suppose $T$ is Lipschitz. Then $T$ is bounded. Hence $T$ is continuous.

(c) at the moment, I can only think of $T(x) =x^2$ whichs is not linear but continuous everywhere. However, I can't quite deduce which of (a) or (b) fails to hold.

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    You seem to be misunderstanding the question incorrectly. It did not ask "Show that is T is not linear then either a) or b) fail to hold", but rather "Find an example of T non-linear where a) fails to hold. Do the same for b)." Of course you can think of a function which is continuous at a single point but not everywhere. And you can think of a non-linear function which is continuous, but not Lipschitz.2012-04-15

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$(a)$ and $(b)$ look right. As for $(c)$:

The function f(x) = \begin{cases} 0 & x < 0 \\ 1 & x \geq 0 \\ \end{cases}

Is non-linear and continuous at $5$ but not continuous everywhere on $\mathbb R$. So this is a counterexample for $(a)$. For $(b)$, Wikipedia gives you $f: \mathbb R \to \mathbb R$, $f(x) = x^2$ as an example of a function that's continuous but not Lipschitz continuous.

To see why $f(x) = x^2$ is not Lipschitz continuous note that "$f : X \to Y$ is Lipschitz" means that there exists a constant $K$ such that for all $x \in X$ we have \frac{|f(x) - f(x^\prime)|}{|x - x^\prime|} < K. So in particular, \frac{d}{dx} f(x) = \lim_{\Delta \to 0} \frac{|f(x) - f(x + \Delta)|}{\Delta} < K. For $f(x) = x^2$ you have $f^\prime (x) = 2x$. We are on $\mathbb R$ so this means that for every constant $K$ you can find an $x_0$ such that $f(x_0) > K$ so $f$ cannot be Lipschitz.

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    @kuku I'm glad. : )2012-04-15