I'm studying for a Real Analysis prelim and have the following problem:

"Let $X$ and $Y$ be normed spaces over $\mathbb{R}$ and let $$\mathcal{L}(X, Y) = \{T: X \rightarrow Y \mid T \text{ is bounded and linear}\}.$$ Let $X^{*}$ and $Y^{*}$ be the dual spaces of $X$ and $Y$ respectively, and let $T^{+}: Y^{*} \rightarrow X^{*}$ be defined by $T^{+}f = f \circ T$. Prove that $T^{+}$ is injective if and only if the image of $T$ is dense in $Y$."

I think I have a proof for the forward direction but for some reason the reverse direction is evading me. Please let me know if what I have so far is correct, and if anyone can give a hint or something about the reverse direction, it would be greatly appreciated.

($\Rightarrow$) Suppose $T^{+}$ is injective but that $T(X)$ is not dense in $Y$. Then for some $y \in Y \backslash T(X)$, there exists an $\varepsilon > 0$ such that $||Tx - y||\geq \varepsilon$ for all $x \in X$. Let $f \in Y^{*}$ be defined by $f(z) = || z - y ||$ if $z \in Y\backslash B_{\varepsilon}(y)$ and $\varepsilon$ otherwise. Also, let $g \in Y^{*}$ be defined by $g(z) = ||z - y ||$ for all $z \in Y$. Then $T^{+}f = T^{+}g$ and $f(z) = g(z)$ when $z \in T(X)$ but not always when $z \in Y\backslash T(X)$, contradicting $T^{+}$ being injective.

($\Leftarrow$) For this direction, I'm thinking about showing through contraposition by assuming $T^{+}f = T^{+}g$ with $f(z) \neq g(z)$ for some $z \in Y\backslash T(X)$ and arriving at the conclusion that $T(X)$ is not dense in $Y$, but for now can't think of how to do that.

P.S. This is my first post on StackExchange, so go easy on me if I made a complete fool of myself; Functional Analysis is quite new to me.

  • $\begingroup$ I think you are on the right track. If $f$ and $g$ are linear functionals, they are continuous (I think--I too am rather new to functional analysis). Two continuous functionals equal on a dense set must be equal everywhere. $\endgroup$
    – Plutoro
    Jun 26, 2015 at 5:21
  • $\begingroup$ Okay I see yeah I didn't think about that. I guess my $f$ doesn't work in the first part of the proof since it is not continuous and therefore not a bounded linear functional. $\endgroup$ Jun 26, 2015 at 5:44

1 Answer 1


Suppose $T^+f= T^+g$, which implies $f\circ T = g\circ T$. Now $f$ and $g$ are identical on the dense set, since the range of $T$ is dense. So they have to be equal, proving one-one.


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